NCERT Solutions for Exercise 4.1 Class 11 Maths Chapter 4 - Principle of Mathematical Induction

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NCERT Solutions for Exercise 4.1 Class 11 Maths Chapter 4 - Principle of Mathematical Induction

Edited By Vishal kumar | Updated on Nov 03, 2023 10:56 AM IST

NCERT Solutions for Class 11 Maths Chapter 4 - Principle of Mathematical Induction Exercise 4.1- Download Free PDF

NCERT Solutions for Class 11 Maths Chapter 4 – Principle of Mathematical Induction Exercise 4.1- NCERT solutions for exercise 4.1 Class 11 Maths Chapter 4 deals with the principle of mathematical induction and related problems. Exercise 4.1 Class 11 Maths deals with how to prove a given mathematical statement using the principle of mathematical induction. NCERT Solutions for Class 11 Maths chapter 4 exercise 4.1 gives an insight into the steps of proving a given statement using the idea of induction. Solving both example questions and Class 11 Maths chapter 4 exercise 4.1 are important to understand the concepts discussed in the chapter. Before starting the Class 11 Maths chapter 4 exercise 4.1 the NCERT book discusses the concept of deduction in brief and introduces what is induction and how induction is different from deduction. Followed by examples and Class 11 Maths chapter 4 exercise 4.1.

The NCERT Solutions for Class 11 Maths Exercise 4.1 are expertly crafted to help students overcome challenges in problem-solving. They provide step-by-step explanations and a free PDF version for convenient access, ensuring that students can grasp mathematical concepts effectively and excel in their studies.

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Question:1 Prove the following by using the principle of mathematical induction for all $n\in N$ : $1+3+3^2+...+3^{n-1}=\frac{(3^n-1)}{2}$

Let the given statement be p(n) i.e.
$p(n):1+3+3^2+...+3^{n-1}=\frac{(3^n-1)}{2}$
For n = 1 we have
$p(1): 1=\frac{(3^1-1)}{2}=\frac{3-1}{2}= \frac{2}{2}=1$ , which is true

For n = k we have
$p(k):1+3+3^2+...+3^{k-1}=\frac{(3^k-1)}{2} \ \ \ \ \ \ \ -(i)$ , Let's assume that this statement is true

Now,
For n = k + 1 we have
$p(k+1):1+3+3^2+...+3^{k+1-1}= 1+3+3^2+...+3^{k-1}+3^{k}$
$= (1+3+3^2+...+3^{k-1})+3^{k}$
$= \frac{(3^k-1)}{2}+3^{k} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (Using \ (i))$
$= \frac{3^k-1+2.3^k}{2}$
$= \frac{3^k(1+2)-1}{2}$
$= \frac{3.3^k-1}{2}$
$= \frac{3^{k+1}-1}{2}$
Thus, p(k+1) is true whenever p(k) is true
Hence, by the principle of mathematical induction, statement p(n) is true for all natural numbers n

Let the given statement be p(n) i.e.
$p(n):1^3+2^3+3^3+...+n^3=\left (\frac{n(n+1)}{2} \right )^2$
For n = 1 we have
$p(1):1=\left (\frac{1(1+1)}{2} \right )^2= \left ( \frac{1(2)}{2} \right )^2=(1)^2=1$ , which is true

For n = k we have
$p(k):1^3+2^3+3^3+...+k^3=\left (\frac{k(k+1)}{2} \right )^2 \ \ \ \ \ \ \ \ \ \ - (i)$ , Let's assume that this statement is true

Now,
For n = k + 1 we have
$p(k+1):1^3+2^3+3^3+...+(k+1)^3=1^3+2^3+3^3+...+k^3+(k+1)^3$
$=(1^3+2^3+3^3+...+k^3)+(k+1)^3$
$=\left ( \frac{k(k+1)}{2} \right )^2+(k+1)^3 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (using \ (i))$
$=\frac{k^2(k+1)^2+4(k+1)^3}{4}$
$=\frac{(k+1)^2(k^2+4(k+1))}{4}$
$=\frac{(k+1)^2(k^2+4k+4)}{4}$
$=\frac{(k+1)^2(k+2)^2}{4} \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because a^2+2ab+b^2=(a+b)^2)$
$=\left ( \frac{(k+1)(k+2)}{2} \right )^2$
Thus, p(k+1) is true whenever p(k) is true
Hence, by the principle of mathematical induction, statement p(n) is true for all natural numbers n

Let the given statement be p(n) i.e.
$p(n):1+\frac{1}{(1+2)}+\frac{1}{(1+2+3)}+...+\frac{1}{(1+2+3+...+n)}=\frac{2n}{(n+1)}$
For n = 1 we have
$p(1):1=\left (\frac{2(1)}{1+1} \right )= \left ( \frac{2}{2} \right )=1$ , which is true

For n = k we have
$p(k):1+\frac{1}{(1+2)}+\frac{1}{(1+2+3)}+...+\frac{1}{(1+2+3+...+k)}=\frac{2k}{(k+1)} \ \ \ \ -(i)$ , Let's assume that this statement is true

Now,
For n = k + 1 we have
$p(k+1):1+\frac{1}{(1+2)}+\frac{1}{(1+2+3)}+...+\frac{1}{(1+2+3+...+k+1)}$ $=\left ( 1+\frac{1}{(1+2)}+\frac{1}{(1+2+3)}+...+\frac{1}{(1+2+3+...+k)} \right )+\frac{1}{(1+2+3+...+k+k+1)}$

$=\frac{2k}{k+1}+\frac{1}{(1+2+3+...+k+(k+1))} \ \ \ \ \ \ \ \ \ \ \ \ \ \ (using \ (i))$
$=\frac{2k}{k+1}+\frac{1}{\frac{(k+1)(k+1+1)}{2}} \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because1+2+....+n = \frac{n(n+1)}{2} )$
$=\frac{2k}{k+1}+\frac{2}{(k+1)(k+2)}$
$=\frac{2}{k+1}\left (k+\frac{1}{k+2} \right )$
$=\frac{2}{k+1}\left ( \frac{k^2+2k+1}{k+2} \right )$
$=\frac{2}{k+1}.\frac{(k+1)^2}{k+2}$
$=\frac{2(k+1)}{k+2}$
Thus, p(k+1) is true whenever p(k) is true
Hence, by the principle of mathematical induction, statement p(n) is true for all natural numbers n

Let the given statement be p(n) i.e.
$p(n):1.2.3+2.3.4+...+n(n+1)(n+2)=\frac{n(n+1)(n+2)(n+3)}{4}$
For n = 1 we have
$p(1):6=\left (\frac{1(1+1)(1+2)(1+3)}{4} \right )= \left ( \frac{1.2.3.4}{4} \right )=6$ , which is true

For n = k we have
$p(k):1.2.3+2.3.4+...+k(k+1)(k+2)=\frac{k(k+1)(k+2)(k+3)}{4} \ \ \ \ \ \ -(i)$ , Let's assume that this statement is true

Now,
For n = k + 1 we have
$p(k+1):1.2.3+2.3.4+...+k(k+1)(k+2) + (k+1)(k+2)(k+3)$ $=(1.2.3+2.3.4+...+k(k+1)(k+2)) + (k+1)(k+2)(k+3)$

$=\frac{k(k+1)(k+2)(k+3)}{4} + (k+1)(k+2)(k+3) \ \ \ \ \ \ \ \ \ \ \ \ (using \ (i))$
$=\frac{k(k+1)(k+2)(k+3)+4(k+1)(k+2)(k+3) }{4}$
$=\frac{(k+1)(k+2)(k+3)(k+4) }{4}$
Thus, p(k+1) is true whenever p(k) is true
Hence, by the principle of mathematical induction, statement p(n) is true for all natural numbers n

Let the given statement be p(n) i.e.
$p(n):1.3+2.3^2+3.3^3+...+n.3^n=\frac{(2n-1)3^{n+1}+3}{4}$
For n = 1 we have
$p(1):3=\frac{(2(1)-1)3^{1+1}+3}{4}= \frac{(2-1)9+3}{4}=\frac{12}{4}=3$ , which is true

For n = k we have

$p(k):1.3+2.3^2+3.3^3+...+k.3^k=\frac{(2k-1)3^{k+1}+3}{4} \ \ \ \ \ \ -(i)$ , Let's assume that this statement is true
Now,
For n = k + 1 we have
$p(k+1):1.3+2.3^2+3.3^3+...+(k+1).3^{(k+1)}$$=1.3+2.3^2+3.3^3+...+k.3^k+(k+1).3^{(k+1)}$

$=\frac{(2k-1)3^{k+1}+3}{4}+(k+1).3^{(k+1)}$
$=\frac{(2k-1)3^{k+1}+3+4(k+1).3^{(k+1)}}{4}$
$=\frac{3^{k+1}((2k-1)+4(k+1))+3}{4}$
$=\frac{3^{k+1}(6k+3)+3}{4}$
$=\frac{3^{k+1}.3(2k+1)+3}{4}$
$=\frac{(2k+1)3^{k+2}+3}{4}$
$=\frac{(2(k+1)-1)3^{(k+1)+1}+3}{4}$
Thus, p(k+1) is true whenever p(k) is true
Hence, by the principle of mathematical induction, statement p(n) is true for all natural numbers n

Let the given statement be p(n) i.e.
$p(n):1.2+2.3+3.4+...+n.(n+1)=\left [\frac{n(n+1)(n+2)}{3} \right ]$
For n = 1 we have
$p(1):2=\left [\frac{1(1+1)(1+2)}{3} \right ]= \frac{1.2.3}{3}=2$ , which is true

For n = k we have

$p(k):1.2+2.3+3.4+...+k.(k+1)=\left [\frac{k(k+1)(k+2)}{3} \right ] \ \ \ \ \ \ \ -(i)$ , Let's assume that this statement is true
Now,
For n = k + 1 we have
$p(k+1):1.2+2.3+3.4+...+(k+1).(k+2)$$=1.2+2.3+3.4+...+k(k+1)+(k+1).(k+2)$

$=\frac{k(k+1)(k+2)}{3}+(k+1).(k+2) \ \ \ \ \ \ \ \ \ (using \ (i))$
$=\frac{k(k+1)(k+2)+3(k+1).(k+2)}{3}$
$=\frac{(k+1)(k+2)(k+3)}{3}$
$=\frac{(k+1)(k+1+1)(k+1+2)}{3}$

Thus, p(k+1) is true whenever p(k) is true
Hence, by principle of mathematical induction , statement p(n) is true for all natural numbers n

Let the given statement be p(n) i.e.
$p(n):1.3+3.5+5.7+...+(2n-1)(2n+1)=\frac{n(4n^2+6n-1)}{3}$
For n = 1 we have
$p(1):1.3=3=\frac{1(4(1)^2+6(1)-1)}{3}= \frac{4+6-1}{3}=\frac{9}{3}=3$ , which is true

For n = k we have

$p(k):1.3+3.5+5.7+...+(2k-1)(2k+1)=\frac{k(4k^2+6k-1)}{3} \ \ \ \ \ \ \ \ -(i)$ , Let's assume that this statement is true
Now,
For n = k + 1 we have
$p(k+1):1.3+3.5+5.7+...+(2(k+1)-1)(2(k+1)+1)$$=1.3+3.5+5.7+...+(2k-1)(2k+1)+(2(k+1)-1)(2(k+1)+1)$

$=\frac{k(4k^2+6k-1)}{3}+(2k+1)(2k+3) \ \ \ \ \ \ \ \ \ \ (using \ (i))$
$=\frac{k(4k^2+6k-1)+3(2k+1)(2k+3)}{3}$
$=\frac{k(4k^2+6k-1)+3(4k^2+8k+3)}{3}$
$=\frac{(4k^3+6k^2-k+12k^2+28k+9)}{3}$
$=\frac{(4k^3+18k^2+23k+9)}{3}$
$=\frac{(4k^3+14k^2+9k+4k^2+14k+9)}{3}$
$=\frac{(k(4k^2+14k+9)+4k^2+14k+9)}{3}$
$=\frac{(4k^2+14k+9)(k+1)}{3}$
$=\frac{(k+1)(4k^2+8k+4+6k+6-1)}{3}$
$=\frac{(k+1)(4(k^2+2k+1)+6(k+1)-1)}{3}$
$=\frac{(k+1)(4(k+1)^2+6(k+1)-1)}{3}$

Thus, p(k+1) is true whenever p(k) is true
Hence, by the principle of mathematical induction, statement p(n) is true for all natural numbers n

Let the given statement be p(n) i.e.

For n = 1 we have , which is true

For n = k we have

, Let's assume that this statement is true
Now,
For n = k + 1 we have
$p(k+1):1.2+2.2^2+3.2^3+...+(k+1).2^{k+1}$ $=1.2+2.2^2+3.2^3+...+k.2^k+(k+1).2^{k+1}$

$=(k-1)2^{k+1}+2+(k+1).2^{k+1} \ \ \ \ \ \ \ \ \ \ (using \ (i))$
$=2^{k+1}(k-1+k+1)+2$
$=2^{k+1}(2k)+2$
$=k.2^{k+2}+2$
$=(k+1-1).2^{k+1+1}+2$

Thus, p(k+1) is true whenever p(k) is true
Hence, by the principle of mathematical induction, statement p(n) is true for all natural numbers n

Let the given statement be p(n) i.e.
$p(n):\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2^n}=1-\frac{1}{2^n}$
For n = 1 we have
$p(1):\frac{1}{2}=1-\frac{1}{2^1}= 1-\frac{1}{2} = \frac{1}{2}$ , which is true

For n = k we have

$p(k):\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2^k}=1-\frac{1}{2^k} \ \ \ \ \ \ \ \ \ -(i)$ , Let's assume that this statement is true
Now,
For n = k + 1 we have
$p(k+1):\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2^{k+1}}$$=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2^k}+\frac{1}{2^{k+1}}$

$=1-\frac{1}{2^k}+\frac{1}{2^{k+1}} \ \ \ \ \ \ \ \ \ \ (using \ (i))$
$=1-\frac{1}{2^k}\left (1-\frac{1}{2} \right )$
$=1-\frac{1}{2^k}\left (\frac{1}{2} \right )$
$=1-\frac{1}{2^{k+1}}$

Thus, p(k+1) is true whenever p(k) is true
Hence, by the principle of mathematical induction, statement p(n) is true for all natural numbers n

Let the given statement be p(n) i.e.
$p(n):\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{(3n-1)(3n+2)}=\frac{n}{(6n+4)}$
For n = 1 we have
$p(1):\frac{1}{2.5}= \frac{1}{10}=\frac{1}{(6(1)+4)}= \frac{1}{10}$ , which is true

For n = k we have

$p(k):\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{(3k-1)(3k+2)}=\frac{k}{(6k+4)} \ \ \ \ \ \ \ \ \ -(i)$ , Let's assume that this statement is true
Now,
For n = k + 1 we have
$p(k+1):\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{(3(k+1)-1)(3(k+1)+2)}$$=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{(3k-1)(3k+2)}+\frac{1}{(3(k+1)-1)(3(k+1)+2)}$

$=\frac{k}{6k+4}+\frac{1}{(3k+2)(3k+5)} \ \ \ \ \ \ \ \ \ \ (using \ (i))$
$=\frac{1}{3k+2}\left ( \frac{k}{2}+\frac{1}{3k+5} \right )$
$=\frac{1}{3k+2}\left ( \frac{k(3k+5)+2}{2(3k+5)} \right )$
$=\frac{1}{3k+2}\left ( \frac{3k^2+5k+2}{2(3k+5)} \right )$
$=\frac{1}{3k+2}\left ( \frac{3k^2+3k+2k+2}{2(3k+5)} \right )$
$=\frac{1}{3k+2}\left ( \frac{(3k+2)(k+1)}{2(3k+5)} \right )$
$=\frac{(k+1)}{6k+10}$
$=\frac{(k+1)}{6(k+1)+4}$

Thus, p(k+1) is true whenever p(k) is true
Hence, by the principle of mathematical induction, statement p(n) is true for all natural numbers n

Let the given statement be p(n) i.e.
$p(n):\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{n(n+1)(n+2)}=\frac{n(n+3)}{4(n+1)(n+2)}$
For n = 1 we have
$p(1):\frac{1}{1.2.3}=\frac{1}{6}=\frac{1(1+3)}{4(1+1)(1+2)}=\frac{4}{4.2.3}=\frac{1}{6}$ , which is true

For n = k we have

$p(k):\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{k(k+1)(k+2)}=\frac{k(k+3)}{4(k+1)(k+2)} \ \ \ \ -(i)$ , Let's assume that this statement is true
Now,
For n = k + 1 we have
$p(k+1):\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{(k+1)(k+2)(k+3)}$$=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{k(k+1)(k+2)}+\frac{1}{(k+1)(k+2)(k+3)}$

$=\frac{k(k+3)}{4(k+1)(k+2)}+\frac{1}{(k+1)(k+2)(k+3)} \ \ \ \ \ \ (using \ (i))$
$=\frac{1}{(k+1)(k+2)}\left ( \frac{k(k+3)}{4}+ \frac{1}{k+3} \right )$
$=\frac{1}{(k+1)(k+2)}\left ( \frac{k(k+3)^2+4}{4(k+3)} \right )$
$=\frac{1}{(k+1)(k+2)}\left ( \frac{k(k^2+9+6k)+4}{4(k+3)} \right )$
$=\frac{1}{(k+1)(k+2)}\left ( \frac{k^3+9k+6k^2+4}{4(k+3)} \right )$
$=\frac{1}{(k+1)(k+2)}\left ( \frac{k^3+2k^2+k+8k+4k^2+4}{4(k+3)} \right )$
$=\frac{1}{(k+1)(k+2)}\left ( \frac{k(k^2+2k+1)+4(k^2+2k+1)}{4(k+3)} \right )$
$=\frac{1}{(k+1)(k+2)}\left ( \frac{(k+1)^2(k+4)}{4(k+3)} \right )$
$= \frac{(k+1)((k+1)+3)}{4(k+1+1)(k+1+2)}$

Thus, p(k+1) is true whenever p(k) is true
Hence, by the principle of mathematical induction, statement p(n) is true for all natural numbers n

Let the given statement be p(n) i.e.
$p(n):a+ar+ar^2+...+ar^{n-1}=\frac{a(r^n-1)}{r-1}$
For n = 1 we have
$p(1):a=\frac{a(r^1-1)}{r-1}=\frac{r-1}{r-1}=1$ , which is true

For n = k we have

$p(k):a+ar+ar^2+...+ar^{k-1}=\frac{a(r^k-1)}{r-1} \ \ \ \ \ \ \ -(i)$ , Let's assume that this statement is true
Now,
For n = k + 1 we have
$p(k+1):a+ar+ar^2+...+ar^{k}$$=a+ar+ar^2+...+ar^{k-1}+ar^{k}$

$=a.\frac{r^k-1}{r-1}+ar^{k} \ \ \ \ \ \ \ \ \ \ \ \ \ (using \ (i))$
$=\frac{a(r^k-1)+(r-1)ar^{k}}{r-1}$
$=\frac{ar^k(1+r-1)-a}{r-1}$
$=\frac{ar^k.r-a}{r-1}$
$=\frac{a(r^{k+1}-1)}{r-1}$

Thus, p(k+1) is true whenever p(k) is true
Hence, by the principle of mathematical induction, statement p(n) is true for all natural numbers n

Let the given statement be p(n) i.e.
$p(n):\left ( 1+\frac{3}{1} \right )\left ( 1+\frac{5}{4} \right )\left ( 1+\frac{7}{9} \right )..\left ( 1+\frac{(2n+1)}{n^2} \right )=(n+1)^2$
For n = 1 we have
$p(1):\left ( 1+\frac{3}{1} \right )= 4=(1+1)^2=2^2=4$ , which is true

For n = k we have

$p(k):\left ( 1+\frac{3}{1} \right )\left ( 1+\frac{5}{4} \right )\left ( 1+\frac{7}{9} \right )..\left ( 1+\frac{(2k+1)}{k^2} \right )=(k+1)^2 \ \ \ \ \ \ \ \ -(i)$ , Let's assume that this statement is true
Now,
For n = k + 1 we have
$p(k+1):\left ( 1+\frac{3}{1} \right )\left ( 1+\frac{5}{4} \right )\left ( 1+\frac{7}{9} \right )..\left ( 1+\frac{(2(k+1)+1)}{(k+1)^2} \right )$ $=\left ( 1+\frac{3}{1} \right )\left ( 1+\frac{5}{4} \right )\left ( 1+\frac{7}{9} \right )..\left ( 1+\frac{2k+1}{k^2} \right )\left ( 1+\frac{(2(k+1)+1)}{(k+1)^2} \right )$

$=(k+1)^2\left ( 1+\frac{(2(k+1)+1)}{(k+1)^2} \right ) \ \ \ \ \ \ \ \ \ \ (using \ (i))$
$=(k+1)^2\left ( \frac{{}(k+1)^2+(2(k+1)+1)}{(k+1)^2} \right )$
$=(k^2+1+2k+2k+2+1)$
$=(k^2+4k+4)$
$=(k+2)^2$
$=(k+1+1)^2$

Thus, p(k+1) is true whenever p(k) is true
Hence, by the principle of mathematical induction, statement p(n) is true for all natural numbers n

Let the given statement be p(n) i.e.
$p(n):\left ( 1+\frac{1}{1} \right )\left ( 1+\frac{1}{2} \right )\left ( 1+\frac{1}{3} \right )...\left ( 1+\frac{1}{n} \right )=(n+1)$
For n = 1 we have
$p(1):\left ( 1+\frac{1}{1} \right )=2=(1+1)=2$ , which is true

For n = k we have

$p(k):\left ( 1+\frac{1}{1} \right )\left ( 1+\frac{1}{2} \right )\left ( 1+\frac{1}{3} \right )...\left ( 1+\frac{1}{k} \right )=(k+1) \ \ \ \ \ \ \ -(i)$ , Let's assume that this statement is true
Now,
For n = k + 1 we have
$p(k+1):\left ( 1+\frac{1}{1} \right )\left ( 1+\frac{1}{2} \right )\left ( 1+\frac{1}{3} \right )...\left ( 1+\frac{1}{k+1} \right )$ &nbsnbsp; $=\left ( 1+\frac{1}{1} \right )\left ( 1+\frac{1}{2} \right )\left ( 1+\frac{1}{3} \right )...\left ( 1+\frac{1}{k} \right )\left ( 1+\frac{1}{k+1} \right )$

$=(k+1)\left ( 1+\frac{1}{k+1} \right ) \ \ \ \ \ \ \ (using \ (i))$
$=(k+1)\left ( \frac{k+1+1}{k+1} \right )$
$=(k+2)$
$=(k+1+1)$
Thus, p(k+1) is true whenever p(k) is true
Hence, by the principle of mathematical induction, statement p(n) is true for all natural numbers n

Let the given statement be p(n) i.e.
$p(n):1^2+3^2+5^2+...+(2n-1)^2=\frac{n(2n-1)(2n+1)}{3}$
For n = 1 we have
$p(1):1^2=1=\frac{1(2(1)-1)(2(1)+1)}{3}= \frac{1.1.3}{3}=1$ , which is true

For n = k we have

$p(k):1^2+3^2+5^2+...+(2k-1)^2=\frac{k(2k-1)(2k+1)}{3} \ \ \ \ \ \ \ \ \ \ -(i)$ , Let's assume that this statement is true
Now,
For n = k + 1 we have
$p(k+1):1^2+3^2+5^2+...+(2(k+1)-1)^2$$=1^2+3^2+5^2+...+(2k-1)^2+(2(k+1)-1)^2$

$=\frac{k(2k-1)(2k+1)}{3}+(2(k+1)-1)^2 \ \ \ \ \ \ \ \ \ \ (using \ (i))$
$=\frac{k(2k-1)(2k+1)+3(2(k+1)-1)^2}{3}$
$=\frac{k(2k-1)(2k+1)+3(2k+1)^2}{3}$
$=\frac{(2k+1)(k(2k-1)+3(2k+1))}{3}$
$=\frac{(2k+1)(2k^2-k+6k+3)}{3}$
$=\frac{(2k+1)(2k^2+5k+3)}{3}$
$=\frac{(2k+1)(2k^2+2k+3k+3)}{3}$
$=\frac{(2k+1)(2k+3)(k+1)}{3}$
$=\frac{(k+1)(2(k+1)-1)(2(k+1)+1)}{3}$

Thus, p(k+1) is true whenever p(k) is true
Hence, by the principle of mathematical induction, statement p(n) is true for all natural numbers n

Let the given statement be p(n) i.e.
$p(n):\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{(3n-2)(3n+1)}=\frac{n}{(3n+1)}$
For n = 1 we have
$p(1):\frac{1}{1.4}=\frac{1}{4}=\frac{1}{(3(1)+1)}=\frac{1}{3+1}=\frac{1}{4}$ , which is true

For n = k we have

$p(k):\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{(3k-2)(3k+1)}=\frac{k}{(3k+1)} \ \ \ \ \ \ \ -(i)$ , Let's assume that this statement is true
Now,
For n = k + 1 we have
$p(k+1):\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{(3(k+1)-2)(3(k+1)+1)}$$=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{(3k-2)(3k+1)}+\frac{1}{(3(k+1)-2)(3(k+1)+1)}$

$=\frac{k}{3k+1}+\frac{1}{(3k+1)(3k+4)} \ \ \ \ \ \ \ \ \ (using \ (i))$
$=\frac{1}{3k+1}\left ( k+\frac{1}{3k+4} \right )$
$=\frac{1}{3k+1}\left ( \frac{k(3k+4)+1}{3k+4} \right )$
$=\frac{1}{3k+1}\left ( \frac{3k^2+4k+1}{3k+4} \right )$
$=\frac{1}{3k+1}\left ( \frac{3k^2+3k+k+1}{3k+4} \right )$
$=\frac{1}{3k+1}\left ( \frac{(3k+1)(k+1)}{3k+4} \right )$
$= \frac{(k+1)}{3k+4}$
$= \frac{(k+1)}{3(k+1)+1}$

Thus, p(k+1) is true whenever p(k) is true
Hence, by the principle of mathematical induction, statement p(n) is true for all natural numbers n

Let the given statement be p(n) i.e.
$p(n):\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{(2n+1)(2n+3)}=\frac{n}{3(2n+3)}$
For n = 1 we have
$p(1):\frac{1}{3.5}= \frac{1}{15}=\frac{1}{3(2(1)+3)}=\frac{1}{3.5}=\frac{1}{15}$ , which is true

For n = k we have

$p(k):\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{(2k+1)(2k+3)}=\frac{k}{3(2k+3)} \ \ \ \ \ \ \ \ \ \ -(i)$ , Let's assume that this statement is true
Now,
For n = k + 1 we have
$p(k+1):\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{(2(k+1)+1)(2(k+1)+3)}$$=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{(2k+1)(2k+3)}+\frac{1}{(2(k+1)+1)(2(k+1)+3)}$

$=\frac{k}{3(2k+3)}+\frac{1}{(2k+3)(2k+5)} \ \ \ \ \ \ \ \ \ \ (using \ (i))$
$=\frac{1}{2k+3}\left ( \frac{k}{3}+\frac{1}{2k+5} \right )$
$=\frac{1}{2k+3}\left ( \frac{k(2k+5)+3}{3(2k+5)} \right )$
$=\frac{1}{2k+3}\left ( \frac{2k^2+5k+3}{3(2k+5)} \right )$
$=\frac{1}{2k+3}\left ( \frac{2k^2+2k+3k+3}{3(2k+5)} \right )$
$=\frac{1}{2k+3}\left ( \frac{(2k+3)(k+1)}{3(2k+5)} \right )$
$= \frac{(k+1)}{3(2k+5)}$
$= \frac{(k+1)}{3(2(k+1)+3)}$

Thus, p(k+1) is true whenever p(k) is true
Hence, by the principle of mathematical induction, statement p(n) is true for all natural numbers n

Let the given statement be p(n) i.e.
$p(n):1+2+3+...+n<\frac{1}{8}(2n+1)^2.$
For n = 1 we have
$p(1):1<\frac{1}{8}(2(1)+1)^2= \frac{1}{8}(3)^2=\frac{9}{8}$ , which is true

For n = k we have

$p(k):1+2+3+...+k<\frac{1}{8}(2k+1)^2 \ \ \ \ \ \ \ \ \ -(i)$ , Let's assume that this statement is true
Now,
For n = k + 1 we have
$p(k+1):1+2+3+...+k+1$$=1+2+3+...+k+k+1$

$< \frac{1}{8}\left ( 2k+1 \right )^2+(k+1) \ \ \ \ \ \ \ \ \ (using \ (i))$
$< \frac{1}{8}\left ( (2k+1)^2+8(k+1) \right )$
$< \frac{1}{8}\left ( 4k^2+4k+1+8k+8 \right )$
$< \frac{1}{8}\left ( 4k^2+12k+9\right )$
$< \frac{1}{8}\left ( 2k+3\right )^2$
$< \frac{1}{8}\left ( 2(k+1)+1\right )^2$

Thus, p(k+1) is true whenever p(k) is true
Hence, by the principle of mathematical induction, statement p(n) is true for all natural numbers n

Let the given statement be p(n) i.e.
$p(n):n(n+1)(n+5)$
For n = 1 we have
$p(1):1(1+1)(1+5)=1.2.6=12$ , which is multiple of 3, hence true

For n = k we have

$p(k):k(k+1)(k+5) \ \ \ \ \ \ \ -(i)$ , Let's assume that this is multiple of 3 = 3m
Now,
For n = k + 1 we have
$p(k+1):(k+1)((k+1)+1)((k+1)+5)$$=(k+1)(k+2)((k+5)+1)$

$=(k+1)(k+2)(k+5)+(k+1)(k+2)$
$=k(k+1)(k+5)+2(k+1)(k+5)+(k+1)(k+2)$
$=3m+2(k+1)(k+5)+(k+1)(k+2) \ \ \ \ \ \ \ \ (using \ (i))$
$=3m+(k+1)(2(k+5)+(k+2)) \ \ \ \ \ \ \ \ (using \ (i))$
$=3m+(k+1)(2k+10+k+2)$
$=3m+(k+1)(3k+12)$
$=3m+3(k+1)(k+4)$
$=3(m+(k+1)(k+4) )$
$=3l$ Where $\left ( l=(m+(k+1)(k+4) ) \right )$ some natural number

Thus, p(k+1) is true whenever p(k) is true
Hence, by the principle of mathematical induction, statement p(n) is multiple of 3 for all natural numbers n

Let the given statement be p(n) i.e.
$p(n):10^{2n-1}+1$
For n = 1 we have
$p(1):10^{2(1)-1}+1= 10^{2-1}+1=10^1+1=11$ , which is divisible by 11, hence true

For n = k we have

$p(k):10^{2k-1}+1 \ \ \ \ \ \ \ \ \ \ \ -(i)$ , Let's assume that this is divisible by 11 = 11m
Now,
For n = k + 1 we have
$p(k+1):10^{2(k+1)-1}+1$$=10^{2k+2-1}+1$
nbsp;
$=10^{2k+1}+1$
$=10^2(10^{2k-1}+1-1)+1$
$=10^2(10^{2k-1}+1)-10^2+1$
$=10^2(11m)-100+1 \ \ \ \ \ \ \ \ \ \ \ \ (using \ (i))$
$=100(11m)-99$
$=11(100m-9)$
$=11l$ Where $l= (100m-9)$ some natural number

Thus, p(k+1) is true whenever p(k) is true
Hence, by the principle of mathematical induction, statement p(n) is divisible by 11 for all natural numbers n

Let the given statement be p(n) i.e.
$p(n):x^2^n-y^2^n$
For n = 1 we have
$p(1):x^{2(1)}-y^{2(1)}= x^2-y^2=(x-y)(x+y)$ , which is divisible by $(x+y)$ , hence true $(using \ a^2-b^2=(a+b)(a-b))$

For n = k we have

$p(k):x^{2k}-y^{2k} \ \ \ \ \ \ \ \ \ \ -(i)$ , Let's assume that this is divisible by $(x+y)$ $=(x+y)m$
Now,
For n = k + 1 we have
$p(k+1):x^{2(k+1)}-y^{2(k+1)}$ $=x^{2k}.x^2-y^{2k}.y^2$

$=x^2(x^{2k}+y^{2k}-y^{2k})-y^{2k}.y^2$
$=x^2(x^{2k}-y^{2k})+x^2.y^{2k}-y^{2k}.y^2$
$=x^2(x+y)m+(x^2-y^2)y^{2k} \ \ \ \ \ \ \ \ \ \ \ \ (using \ (i))$
$=x^2(x+y)m+((x-y)(x+y))y^{2k} \ \ \ \ \ \ \ \ \ \ \ \ (using \ a^2-b^2=(a+b)(a-b))$
$=(x+y)\left ( x^2.m+(x-y).y^{2k} \right )$
$=(x+y)l$ where $l = (x^2.m+(x-y).y^{2k})$ some natural number

Thus, p(k+1) is true whenever p(k) is true
Hence, by the principle of mathematical induction, statement p(n) is divisible by $(x+y)$ for all natural numbers n

Let the given statement be p(n) i.e.
$p(n):3^{2n+2}-8n-9$
For n = 1 we have
$p(1):3^{2(1)+2}-8(1)-9= 3^4-8-9=81-17=64=8\times 8$ , which is divisible by 8, hence true

For n = k we have

$p(k):3^{2k+2}-8k-9 \ \ \ \ \ \ \ \ \ -(i)$ , Let's assume that this is divisible by 8 = 8m
Now,
For n = k + 1 we have
$p(k+1):3^{2(k+1)+2}-8(k+1)-9$$=3^{2k+2+2}-8(k+1)-9$
$=3^{2k+2}.3^2-8k-8-9$
$=3^2(3^{2k+2}-8k-9+8k+9)-8k-17$
$=3^2(3^{2k+2}-8k-9)+3^2(8k+9)-8k-17$
$=9\times 8m+9(8k+9)-8k-17 \ \ \ \ \ \ \ \ \ \ \ \ (using \ (i))$
$=9\times 8m+72k+81-8k-17$
$=9\times 8m+80k-64$
$=9\times 8m+8(10k-8)$
$=8(9m+10k-8)$
$=8l$ where $l= 9m+10k-8$ some natural number

Thus, p(k+1) is true whenever p(k) is true
Hence, by the principle of mathematical induction, statement p(n) is divisible by 8 for all natural numbers n

Let the given statement be p(n) i.e.
$p(n):41^n-14^n$
For n = 1 we have
$p(1):41^1-14^1= 41-14= 27$ , which is divisible by 27, hence true

For n = k we have

$p(k):41^k-14^k \ \ \ \ \ \ \ \ \ \ \ -(i)$ , Let's assume that this is divisible by 27 = 27m
Now,
For n = k + 1 we have
$p(k+1):41^{k+1}-14^{k+1}$$=41^{k}.41-14^{k}.14$
$=41(41^{k}-14^k+14^k)-14^{k}.14$
$=41(41^{k}-14^k)+14^k.41-14^{k}.14$
$=41(27m)+14^k(41-14) \ \ \ \ \ \ \ \ \ \ \ (using \ (i))$
$=41(27m)+14^k.27$
$=27(41m+14^k)$
$=27l$ where $l = 41m+14^k$ some natural number

Thus, p(k+1) is true whenever p(k) is true
Hence, by the principle of mathematical induction, statement p(n) is divisible by 27 for all natural numbers n

Let the given statement be p(n) i.e.
$p(n):(2n+7)<(n+3)^2$
For n = 1 we have
$p(1):(2(1)+7)<(1+3)^2\Rightarrow 9< 16$ , which is true

For n = k we have

$p(k):(2k+7)<(k+3)^2 \ \ \ \ \ \ \ \ \ \ \ -(i)$ , Let's assume that this is true
Now,
For n = k + 1 we have
$p(k+1):(2(k+1)+7)$$=(2k+2+7)$
$<(k+3)^2+2 \ \ \ \ \ \ \ \ \ \ \ (using \ (i))$
$
$
$Now , \
$<(k+4)^2$
$<((k+1)+3)^2$

Thus, p(k+1) is true whenever p(k) is true
Hence, by the principle of mathematical induction, statement p(n) is true for all natural numbers n

More About NCERT Solutions for Class 11 Maths Chapter 4 Exercise 4.1

Before the introduction of exercise 4.1 Class 11 Maths 8 example problems are given in NCERT chapter 4 of Class 11. Only one exercise is discussed in this chapter, which is Class 11 Maths chapter 4 exercise 4.1. 24 questions are discussed in the Class 11 Maths chapter 4 exercise 4.1. Solving all these questions are important as students may face similar or same questions in the exam paper. Give a good number of try to solve questions before looking to the NCERT solutions for Class 11 Maths chapter 4 exercise 4.1.

Benefits of NCERT Solutions for Class 11 Maths Chapter 4 Exercise 4.1

• Solving all the questions of exercise 4.1 Class 11 Maths gives a good idea of the steps involved in solving a given problem using mathematical induction

• From this 11th class maths exercise 4.1 answers, students can definitely expect a question for the Class 11 final exam. To solve questions in the exam paper it is important to practice NCERT solutions for Class 11 Maths chapter 4 exercise 4.1.

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Key Features of 11th Class Maths Exercise 4.1 Answers

1. Step-by-step explanations: Detailed ex 4.1 class 11 solutions with a step-by-step breakdown for each problem.

2. Clarity and precision: Clear and accurate presentation, helping students confidently prepare for exams.

3. Conceptual understanding: Emphasis on fostering a deep understanding of mathematical concepts.

4. Curriculum alignment: class 11 maths ex 4.1 solutions closely follow the NCERT curriculum, covering topics as per the official syllabus.

5. Variety of practice problems: A range of class 11 maths chapter 4 exercise 4.1 exercises provided for students to enhance their problem-solving skills.

6. Accessibility: Class 11 ex 4.1 solutions are often available for free, making them easily accessible to students.

7. Format options: PDF versions are typically offered for convenient online and offline access.

Subject Wise NCERT Exampler Solutions

1. How many topics are given in the NCERT Class 11 chapter 4?

Three topics

2. What are the three topics given in the chapter principle of mathematical induction?

Introduction, motivation and the principle of mathematical induction

3. How many questions are solved in the Class 11 Maths chapter 4 exercise 4.1?

24 questions

4. What number of solved examples are given in the NCERT Class 11 Maths Chapter 4 - Principle of Mathematical Induction?

Eight examples are solved before the Exercise 4.1 Class 11 Maths

5. What is the relation between the principle of mathematical induction and Peano’s axioms?

It is a restatement of one of Peano’s axioms

6. Who was the first person to name and define mathematical induction?

De Morgan

7. Which mathematician is credited with the origin of the principle of mathematical induction?

Blaise Pascal

8. Why do we solve NCERT Solutions for Class 11 Maths chapter 4 exercise 4.1?

The basic aim of solving the NCERT exercise is to understand the concepts well and to get an insight in to different types of questions that uses the concepts studied. Other than this Class 11th Maths chapter 4 exercise 4.1 will help in exams also.

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Orthotist and Prosthetist

Orthotists and Prosthetists are professionals who provide aid to patients with disabilities. They fix them to artificial limbs (prosthetics) and help them to regain stability. There are times when people lose their limbs in an accident. In some other occasions, they are born without a limb or orthopaedic impairment. Orthotists and prosthetists play a crucial role in their lives with fixing them to assistive devices and provide mobility.

6 Jobs Available
Pathologist

A career in pathology in India is filled with several responsibilities as it is a medical branch and affects human lives. The demand for pathologists has been increasing over the past few years as people are getting more aware of different diseases. Not only that, but an increase in population and lifestyle changes have also contributed to the increase in a pathologist’s demand. The pathology careers provide an extremely huge number of opportunities and if you want to be a part of the medical field you can consider being a pathologist. If you want to know more about a career in pathology in India then continue reading this article.

5 Jobs Available
Gynaecologist

Gynaecology can be defined as the study of the female body. The job outlook for gynaecology is excellent since there is evergreen demand for one because of their responsibility of dealing with not only women’s health but also fertility and pregnancy issues. Although most women prefer to have a women obstetrician gynaecologist as their doctor, men also explore a career as a gynaecologist and there are ample amounts of male doctors in the field who are gynaecologists and aid women during delivery and childbirth.

4 Jobs Available
Audiologist

The audiologist career involves audiology professionals who are responsible to treat hearing loss and proactively preventing the relevant damage. Individuals who opt for a career as an audiologist use various testing strategies with the aim to determine if someone has a normal sensitivity to sounds or not. After the identification of hearing loss, a hearing doctor is required to determine which sections of the hearing are affected, to what extent they are affected, and where the wound causing the hearing loss is found. As soon as the hearing loss is identified, the patients are provided with recommendations for interventions and rehabilitation such as hearing aids, cochlear implants, and appropriate medical referrals. While audiology is a branch of science that studies and researches hearing, balance, and related disorders.

3 Jobs Available
Oncologist

An oncologist is a specialised doctor responsible for providing medical care to patients diagnosed with cancer. He or she uses several therapies to control the cancer and its effect on the human body such as chemotherapy, immunotherapy, radiation therapy and biopsy. An oncologist designs a treatment plan based on a pathology report after diagnosing the type of cancer and where it is spreading inside the body.

3 Jobs Available
Anatomist

Are you searching for an ‘Anatomist job description’? An Anatomist is a research professional who applies the laws of biological science to determine the ability of bodies of various living organisms including animals and humans to regenerate the damaged or destroyed organs. If you want to know what does an anatomist do, then read the entire article, where we will answer all your questions.

2 Jobs Available
Actor

For an individual who opts for a career as an actor, the primary responsibility is to completely speak to the character he or she is playing and to persuade the crowd that the character is genuine by connecting with them and bringing them into the story. This applies to significant roles and littler parts, as all roles join to make an effective creation. Here in this article, we will discuss how to become an actor in India, actor exams, actor salary in India, and actor jobs.

4 Jobs Available
Acrobat

Individuals who opt for a career as acrobats create and direct original routines for themselves, in addition to developing interpretations of existing routines. The work of circus acrobats can be seen in a variety of performance settings, including circus, reality shows, sports events like the Olympics, movies and commercials. Individuals who opt for a career as acrobats must be prepared to face rejections and intermittent periods of work. The creativity of acrobats may extend to other aspects of the performance. For example, acrobats in the circus may work with gym trainers, celebrities or collaborate with other professionals to enhance such performance elements as costume and or maybe at the teaching end of the career.

3 Jobs Available
Video Game Designer

Career as a video game designer is filled with excitement as well as responsibilities. A video game designer is someone who is involved in the process of creating a game from day one. He or she is responsible for fulfilling duties like designing the character of the game, the several levels involved, plot, art and similar other elements. Individuals who opt for a career as a video game designer may also write the codes for the game using different programming languages.

Depending on the video game designer job description and experience they may also have to lead a team and do the early testing of the game in order to suggest changes and find loopholes.

3 Jobs Available

Radio Jockey is an exciting, promising career and a great challenge for music lovers. If you are really interested in a career as radio jockey, then it is very important for an RJ to have an automatic, fun, and friendly personality. If you want to get a job done in this field, a strong command of the language and a good voice are always good things. Apart from this, in order to be a good radio jockey, you will also listen to good radio jockeys so that you can understand their style and later make your own by practicing.

A career as radio jockey has a lot to offer to deserving candidates. If you want to know more about a career as radio jockey, and how to become a radio jockey then continue reading the article.

3 Jobs Available
Choreographer

The word “choreography" actually comes from Greek words that mean “dance writing." Individuals who opt for a career as a choreographer create and direct original dances, in addition to developing interpretations of existing dances. A Choreographer dances and utilises his or her creativity in other aspects of dance performance. For example, he or she may work with the music director to select music or collaborate with other famous choreographers to enhance such performance elements as lighting, costume and set design.

2 Jobs Available
Social Media Manager

A career as social media manager involves implementing the company’s or brand’s marketing plan across all social media channels. Social media managers help in building or improving a brand’s or a company’s website traffic, build brand awareness, create and implement marketing and brand strategy. Social media managers are key to important social communication as well.

2 Jobs Available
Photographer

Photography is considered both a science and an art, an artistic means of expression in which the camera replaces the pen. In a career as a photographer, an individual is hired to capture the moments of public and private events, such as press conferences or weddings, or may also work inside a studio, where people go to get their picture clicked. Photography is divided into many streams each generating numerous career opportunities in photography. With the boom in advertising, media, and the fashion industry, photography has emerged as a lucrative and thrilling career option for many Indian youths.

2 Jobs Available
Producer

An individual who is pursuing a career as a producer is responsible for managing the business aspects of production. They are involved in each aspect of production from its inception to deception. Famous movie producers review the script, recommend changes and visualise the story.

They are responsible for overseeing the finance involved in the project and distributing the film for broadcasting on various platforms. A career as a producer is quite fulfilling as well as exhaustive in terms of playing different roles in order for a production to be successful. Famous movie producers are responsible for hiring creative and technical personnel on contract basis.

2 Jobs Available
Copy Writer

In a career as a copywriter, one has to consult with the client and understand the brief well. A career as a copywriter has a lot to offer to deserving candidates. Several new mediums of advertising are opening therefore making it a lucrative career choice. Students can pursue various copywriter courses such as Journalism, Advertising, Marketing Management. Here, we have discussed how to become a freelance copywriter, copywriter career path, how to become a copywriter in India, and copywriting career outlook.

5 Jobs Available
Vlogger

In a career as a vlogger, one generally works for himself or herself. However, once an individual has gained viewership there are several brands and companies that approach them for paid collaboration. It is one of those fields where an individual can earn well while following his or her passion.

Ever since internet costs got reduced the viewership for these types of content has increased on a large scale. Therefore, a career as a vlogger has a lot to offer. If you want to know more about the Vlogger eligibility, roles and responsibilities then continue reading the article.

3 Jobs Available
Publisher

For publishing books, newspapers, magazines and digital material, editorial and commercial strategies are set by publishers. Individuals in publishing career paths make choices about the markets their businesses will reach and the type of content that their audience will be served. Individuals in book publisher careers collaborate with editorial staff, designers, authors, and freelance contributors who develop and manage the creation of content.

3 Jobs Available
Journalist

Careers in journalism are filled with excitement as well as responsibilities. One cannot afford to miss out on the details. As it is the small details that provide insights into a story. Depending on those insights a journalist goes about writing a news article. A journalism career can be stressful at times but if you are someone who is passionate about it then it is the right choice for you. If you want to know more about the media field and journalist career then continue reading this article.

3 Jobs Available
Editor

Individuals in the editor career path is an unsung hero of the news industry who polishes the language of the news stories provided by stringers, reporters, copywriters and content writers and also news agencies. Individuals who opt for a career as an editor make it more persuasive, concise and clear for readers. In this article, we will discuss the details of the editor's career path such as how to become an editor in India, editor salary in India and editor skills and qualities.

3 Jobs Available
Reporter

Individuals who opt for a career as a reporter may often be at work on national holidays and festivities. He or she pitches various story ideas and covers news stories in risky situations. Students can pursue a BMC (Bachelor of Mass Communication), B.M.M. (Bachelor of Mass Media), or MAJMC (MA in Journalism and Mass Communication) to become a reporter. While we sit at home reporters travel to locations to collect information that carries a news value.

2 Jobs Available
Corporate Executive

Are you searching for a Corporate Executive job description? A Corporate Executive role comes with administrative duties. He or she provides support to the leadership of the organisation. A Corporate Executive fulfils the business purpose and ensures its financial stability. In this article, we are going to discuss how to become corporate executive.

2 Jobs Available
Multimedia Specialist

A multimedia specialist is a media professional who creates, audio, videos, graphic image files, computer animations for multimedia applications. He or she is responsible for planning, producing, and maintaining websites and applications.

2 Jobs Available
Welding Engineer

Welding Engineer Job Description: A Welding Engineer work involves managing welding projects and supervising welding teams. He or she is responsible for reviewing welding procedures, processes and documentation. A career as Welding Engineer involves conducting failure analyses and causes on welding issues.

5 Jobs Available
4 Jobs Available
Quality Controller

A quality controller plays a crucial role in an organisation. He or she is responsible for performing quality checks on manufactured products. He or she identifies the defects in a product and rejects the product.

A quality controller records detailed information about products with defects and sends it to the supervisor or plant manager to take necessary actions to improve the production process.

3 Jobs Available
Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.

3 Jobs Available

A QA Lead is in charge of the QA Team. The role of QA Lead comes with the responsibility of assessing services and products in order to determine that he or she meets the quality standards. He or she develops, implements and manages test plans.

2 Jobs Available
Structural Engineer

A Structural Engineer designs buildings, bridges, and other related structures. He or she analyzes the structures and makes sure the structures are strong enough to be used by the people. A career as a Structural Engineer requires working in the construction process. It comes under the civil engineering discipline. A Structure Engineer creates structural models with the help of computer-aided design software.

2 Jobs Available
Process Development Engineer

The Process Development Engineers design, implement, manufacture, mine, and other production systems using technical knowledge and expertise in the industry. They use computer modeling software to test technologies and machinery. An individual who is opting career as Process Development Engineer is responsible for developing cost-effective and efficient processes. They also monitor the production process and ensure it functions smoothly and efficiently.

2 Jobs Available
4 Jobs Available
AWS Solution Architect

An AWS Solution Architect is someone who specializes in developing and implementing cloud computing systems. He or she has a good understanding of the various aspects of cloud computing and can confidently deploy and manage their systems. He or she troubleshoots the issues and evaluates the risk from the third party.

4 Jobs Available

An Azure Administrator is a professional responsible for implementing, monitoring, and maintaining Azure Solutions. He or she manages cloud infrastructure service instances and various cloud servers as well as sets up public and private cloud systems.

4 Jobs Available
Computer Programmer

Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.

3 Jobs Available
Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.

3 Jobs Available
Information Security Manager

Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack

3 Jobs Available
3 Jobs Available
Automation Test Engineer

An Automation Test Engineer job involves executing automated test scripts. He or she identifies the project’s problems and troubleshoots them. The role involves documenting the defect using management tools. He or she works with the application team in order to resolve any issues arising during the testing process.

2 Jobs Available