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You've always wondered how to prove the truth of a pronouncement for all natural numbers or how to prove that a form embraces indefinitely. The answer lies in the principle of Mathematical Induction, which is explained in Class 11, Chapter 4 of mathematics. Mathematical induction is a technique to prove mathematical results that hold for all natural numbers. This article gives you an introduction to the principle of induction and helps you to understand how it is used as a fundamental tool in mathematics. You will understand how to find the truth of a mathematical statement that seems to be complicated beyond the initial, but can still be proved easily by applying induction. Your ability to solve difficult problems will be greatly improved if you understand the method of inductive proofs, where you prove a result for some values and then assume that the statement remains unchanged for other sets of values.
Regular practice with NCERT Exercises and NCERT Worksheets will help you achieve certainty and sharpen your logical deduction skills, preparing you for joint examinations and practical troubleshooting. Hence, we must be ready to unlock the powers of mathematical initiation and to discover the way in which the tenet can connect and simplify the apparently complex mathematical notions.
Class 11 Maths Chapter 4 exemplar solutions Exercise: 4.3 Page number: 70-72 Total questions: 30 |
Question:1
Answer:
Given:
$P(n)$ which it true for all $n \geq 4$ but $P(1), P(2) \: \: and \: \: P(3)$ are not true.
Let us consider P (n) be $2^n< n!$
Examples of the given statements are-
$P (0) = 2^0<0!$ i.e., $1<1 \rightarrow Not True$
$P (1) = 2^1<1!$ i.e., $2<1 \rightarrow Not True$
$P(2) = 2^2<2!$ i.e., $4<2 \rightarrow$ NotTrue
$P(3) = 2^3<3!$ i.e., $8<6 \rightarrow$ NotTrue
$P(4) = 2^4<4!$ i.e., $16<24 \rightarrow$ True
$\& P(5) = 2^5<5!$ i.e., $32<60 \rightarrow$ True.....&so on.
Question:2
Give an example of a statement P (n) which is true for all n. Justify your answer.
Answer:
P (n) is true for all n …….. (Given)
Now, let P (n) be,
$1+ 2 + 3 + 4 +... + n = n (n+1)/2$
P(0) $=0(0+1)/2 = 0 $
$\rightarrow true\\ P (1) $
$\rightarrow1 = 1(2+2)/2 = 1 $
$\rightarrow true\\ P (2)$
$ \rightarrow 1+2 = 2(2+1)/2 $
$\rightarrow true\\ P (k) $
$\rightarrow 1+2+3+... +k = k (k+1)/2\\ P (k+1) $
$\rightarrow 1+2+3+... +k+1 = k (k+1)/2 + k + 1 = (k+1) (k+2)/2 \rightarrow p (k+1)$
is true for all k
Thus, P (n) is true for all n.
Question:3
Answer:
P (n) = $4^n -1$ is divisible by 3 …….. (Given)
Now, we’ll substitute different values for n,
$P (0) = 4^0 - 1 = 0$, viz. divisible by 3
$P (1) = 4^1 - 1 = 3$, divisible by 3.
$P (2) = 4^2 - 1 = 15$, divisible by 3
Now, let $P (k) = 4^k - 1$, is divisible by 3
Thus, $4^k - 1 = 3x$
We also get that,
$P (k+1) = 4^{k+1} - 1 = 4(3x+1) - 1$
$= 12x + 3$, is divisible by 3
Thus, P (k+1) is also true,
Hence, by mathematical induction,
For each natural number n it is true that, $P (n) = 4n -1$ is divisible by 3
Question:4
$2^{3n} -1$ is divisible by7, for all natural numbers n.
Answer:
Now, we'll substitute different values for n,
$
\begin{aligned}
& \mathrm{P}(0)=2^0-1=0, \text { divisible by } 7 \\
& P(1)=2^3-1=7, \text { divisible by } 7 \\
& P(2)=2^6-1=63 \text {, divisible by } 7 \\
& P(3)=2^9-1=512 \text {, divisibleby } 7
\end{aligned}
$
Now, let us consider,
$
P(k)=2^{3 k}-1 \text { be divisible by } 7
$
Thus, $2^{2 k}-1=7 x$
We also get that,
$
P(k+1)=2^{3(k+1)}-1=2^3(3 x+1)-1=56 x+7, \text { is divisible by } 7
$
Thus, $\mathrm{P}(\mathrm{k}+1)$ is also true,
Hence, by mathematical induction, For each natural number n it is true that,
$
\mathrm{P}(\mathrm{n})=2^{3 \mathrm{n}}-1 \text { is divisible by } 7.
$
Question:5
$n^3 -7n + 3$ is divisible by 3 for all natural numbers n.
Answer:
$n^3 -7n + 3$ is divisible by 3... given
Now, we’ll substitute different values for n,
$P (0) = 0^3 -7 \times 0 + 3 = 3$, is divisible by 3
$P (1) = 1^3 -7 \times 1 + 3 = -3$, is divisible by 3
$P (2) = 2^3 -7 \times 2 + 3 = -3$, is divisible by 3
Now, let us consider,
$P (k) = k^3 -7k + 3$ be divisible by 3
Thus, $k^3 -7k + 3 = 3x$
We also get that,
$P (k+1) = (k+1)^3 -7(k+1) + 3 = k^3 + 3k^2 + 3k +1 -7k -7 + 3$
$= 3x + 3(k^2+k-2)$is divisible by 3
Thus, P (k+1) is also true,
Hence, by mathematical induction,
For each natural no. n it is true that, $P (n) = n^3 -7n + 3$ is divisible by 3.
Question:6
$3^{2n -1}$ is divisible by 8, for all natural numbers n.
Answer:
p(n)= $3^{2n -1}$ is divisible by 8.......(given)
Now, we’ll substitute different values for n,
$p(0) = 3^0 -1 = 0$, is divisible by 8
$P(1) = 3^2 -1 = 8$, is divisible by 8
$P(2) = 3^4 -1 = 80$ , is divisible by 8
Now, let us consider,
$P(k) = 3^{2k} -1$ be divisible by 8
Thus, $P(k+1) = 3^{2(k+1)} -1 = 72x + 8$, is divisible by 8
Hence, by mathematical induction,
For each natural number n, it is true that $P(n) = 3^{2n} -1$ is divisible by 8.
Question:7
For any natural number n,$7^n -2^n$ is divisible by 5.
Answer:
P(n)=$7^n -2^n$ is divisible by 5. ...............(given)
Now, we will substitute different values for n,
$P(0) = 7^0 -2^0 = 0$, is divisible by 5
$P(1) = 7^1 -2^1 = 5$, is divisible by 5
$P(2) = 7^2 -2^2 = 45$, is divisible by 5
$P(3) = 7^3 -2^3 = 335$, is divisible by 5
Now, let us consider that,
$P(k) = 7^k -2^k$ be divisible by 5.
Thus, $7^k -2^k=5x$
We also get,
$P(k+1) = 7^{k+1} -2^{k+1} = (5+2)7^k -2(2)^k$
$= 5(7)^k + 2(5x)$, is divisible by 5
Thus, P(k+1) is true
Hence, by mathematical induction,
For each natural number n it is true that $P(n) = 7^n -2^n$ is divisible by 5.
Question:8
For any natural number n, $x^n -y^n$ is divisible by x -y, where x integers with$x \neq y$.
Answer:
P(n)= $x^n -y^n$ is divisible by x -y, where x integers with$x \neq y$. ........(given)
Now, we’ll substitute different values for n,
$P(0) = x^0 -y^0 = 0$, is divisible by x-y
$P(1) = x -y,$ is divisible by x-y
$P(2)=x^2-y^2=(x+y)(x-y)$, is divisible by $\mathrm{x}-\mathrm{y}$
Now, let us consider,
$P(k)=x^k-y^k$, is divisible by x - y
Thus, $x^k-y^k=a(x-y)$
We also get that
$
\begin{aligned}
& P(k+1)=x^{k+1}-y^{k+1} \\
& =x^k(x-y)+y\left(x^k-y^k\right) \\
& =x^k(x-y)+y a(x-y), \text { is divisible by } \mathrm{x}-\mathrm{y}
\end{aligned}
$
Thus, $P(k+1)$ is true
Hence, by mathematical induction,
For each natural no. n it is true that, $P(n)=x^n-y^n$ is divisible by $\mathrm{x}-\mathrm{y}, \mathrm{x}$ integers with $\mathrm{x} \neq \mathrm{y}$.
Question:9
$n^3 -n$ is divisible by 6, for each natural number n. $n \geq2$
Answer:
P(n)= $n^3 -n$ is divisible by 6
Now, we’ll substitute different values for n,
$P(2) = 2^3 -2 = 6$, is divisible by 6
Now, let us consider,
$P(k) = k^3 -k$ be divisible by 6
Thus, $k^3 -k = 6x$
We also get that,
$P(k+1) =(k+1)^3 -(k+1)$
$= (k+1)(k^2 + 2k + 1 -1)$
$= k^3 + 3k^2 + 2k$
$= 6x + 3k(k+1)$
$= 6x + 3 \times 2y$, $\because k(k+1) \text{ is even}$
$= 6x + 3 \times 2y$ is divisible by 6
Thus, P(k+1) is true
Hence, by mathematical induction,
For each natural number n, it is true that P(n)= $n^3 -n$ is divisible by 6.
Question:10
$n(n^2 + 5)$ is divisible by 6 for each natural number n.
Answer:
P(n)= $n(n^2 + 5)$ is divisible by 6 for each natural number n.
Now, we’ll substitute different values for n,
$P(1) = 1(1^2 + 5) = 6$, is divisible by 6
$P(2) = 2(2^{}2 + 5) = 18$, is divisible by 6
Let us consider,
$P(k) = k(k^2 + 5)$be divisible by 6
Thus, $k(k^2 + 5)=6x$
We also get that,
$P(k+1) = (k+1)[(k+1)^{2} + 5]$
$= (k+1)(k^2 + 2k + 6)$
$= k(k^2 + 5 ) + k(2k+1) + k^2+2k+ 6$
$= 6x + 3k^2 + 3k + 6$ $\because (k^2 + k ) \text{ is even }$
$= 6x + 3\times 2y + 6$, is divisible by 6
Thus, P(k+1) is true
Hence, by mathematical induction,
For each natural number n it is true that, P(n)= $n(n^2 + 5)$ is divisible by 6.
Question:11
$n^2 < 2^n$ for all natural numbers $n \geq 5$.
Answer:
P(n)=$n^2 < 2^n$ for $n \geq 5$.....(given)
Let us consider
$P(k) = k^2< 2^k$ to be true,
Thus, $P(k+1) = (k+1)^2$
$= k^2 + 2k + 1$
$2^{k+1} = 2(2^k) > 2k^2$
Now, since, $n^2> 2n + 1$ for $n\geq 3$
$k^2 + 2k + 1 < 2k^2$
Thus, $(k+1)^2< 2^{k+1}$
Thus, P(k+1) is true
Hence, by mathematical induction,
For each natural no. n2< 2n it is true that, P(n)=$n^2 < 2^n$ for $n \geq 5$.
Question:12
$2n < (n + 2)!$ for all natural number n.
Answer:
$2n < (n + 2)!$ .....(given)
Now, we will substitute different values for n,
$P(0) \rightarrow 0 < 2!$
$P(1) \rightarrow 2 < 3!$
$P(2) \rightarrow 4 < 4!$
$P(3) \rightarrow 6 < 5!$
Now, let us consider,
$P(k) = 2k < (k+2)!$ To be true,
Thus, $P(k+1) = 2(k+1)< [(k+1) + 2]!$
Now, we know that,$[(k+1) + 2]! = (k+3)! = [(k+3) (k+2)(k+1) ... 3 \times 2\times 1]$
But also, $= 2(k+1) \times (k +3)(k+2) ... 3 \times 1 > 2(k+1)$
Thus,$2(k+1) < [(k+1) + 2]!$
Thus, P(k+1) is true.
Hence, by mathematical induction,
For each natural no. n it is true that, P(n) = $2n < (n + 2)!$.
Question:13
Answer:
$
\mathrm{P}(n) \text { is } \sqrt{n}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\ldots .+\frac{1}{\sqrt{n}} ; n \geq 2 .
$
$\mathrm{P}(2)$ is $\sqrt{2}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}=1.414<1.7 .0 \ldots$ thus, itistrue $\mathrm{P}(3)$ is $\sqrt{3}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}=1.732<2.284$, thus itis true Now, letusconsider $P(k)=\sqrt{k}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\ldots+\frac{1}{\sqrt{k}}$, is true Now, wewilladd $\sqrt{k+1}$ on both sides,
Thus,
$
\sqrt{k}+\sqrt{k+1}-\sqrt{k}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\ldots+\frac{1}{\sqrt{k}}+\sqrt{k+1}-\sqrt{k}
$
Thus, $\sqrt{k+1}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\ldots .+\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{k+1}}$
Hence, by mathematical induction,
For each natural no. $n \geq 2, \mathrm{P}(\mathrm{n})$ is $\sqrt{n}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\ldots+\frac{1}{\sqrt{n}}$.
Question:14
$2 + 4 + 6 +...+ 2n = n^2 + n$ for all natural numbers n.
Answer:P(n)=$2 + 4 + 6 +...+ 2n = n^2 + n$
Now, we will substitute different values for n,
$P(0) = 0 = 0^2 + 0$, is true
$P(1) = 2 = 1^2 + 1$, is true
$P(2) = 2+4 = 2^2 + 2$, is true
Now, let us consider,
$P(k) = 2 + 4 + 6 + .... + 2k = k^2 + k$ to be true,
Thus,
P(k+1) is$2 + 4 + 6 + .... + 2k + 2(k+1) = k^2 + k + 2k + 2$
$= (k^2 + 2k + 1) + (k+1)$
$= (k + 1)^2 + (k + 1)$
Thus, P(k+1) is true if P(k) is true
Hence, by mathematical induction,
For each natural number n, it is true that P(n)=$2 + 4 + 6 +...+ 2n = n^2 + n$.
Question:15
$1 + 2 + 2^2 + ... 2^n = 2^{n+1} -1$ for all natural numbers n.
Answer:
P(n) is $1 + 2 + 2^2 + ... 2^n = 2^{n+1} -1$ ..................given
Now, we’ll substitute different values for n,
$P(0) = 1 = 2^{0+1} -1$ , is true
$P(1) = 1 + 2 = 3 = 2^{1+1} -1$ , is true
$P(2) = 1 + 2+ 2^2 = 7 = 2^{2+1} -1$, is true
Now, let us consider,
$P(k) = 1 + 2 + 2^2 + .... 2^k = 2^{k+1} -1$ to be true
Thus,
P(k+1) is $1+2+2^2+....+2^k + 2^{k+1 }= 2^{k+1} -1 + 2^{k+1}$
$= 2 \times 2^{k+1 }-1$
$= 2^{k+1+1 }-1$
Thus, P(k+1) is true if P(k) is true
Hence, by mathematical induction,
For each natural number n it is true that, P(n) is $1 + 2 + 2^2 + ... 2^n = 2^{n+1} -1$.
Question:16
$1 + 5 + 9 + ...+(4n -3) = n (2n -1)$ for all natural number n.
Answer:
P(n)= $1 + 5 + 9 + ...+(4n -3) = n (2n -1)$
Now, we’ll substitute different values for n,
P(1) = 1
= 1(2-1), is true
P(2) = 1+5 = 6
= 2(4-1), is true
Now, let us consider,
$P(k) = 1+5+9+.....+[4k-3] = k(2k-1)$
$P(k+1) = 1+5+9+.....+[4(k+1)-3] = k(2k-1) + 4(k+1)-3$
$= k(2k -1) + 4k + 1\\ (k+1)[2(k+1)-1]$
Thus, P(k+1) is true if P(k) is true
Hence, by mathematical induction,
For each natural no. n it is true that, P(n)= $1 + 5 + 9 + ...+(4n -3) = n (2n -1)$
Question:17
Answer:
Given
$
a_1=3 \& a_k=7 a_{k-1}
$
Now, we'll substitute different values for $k$
$
\begin{aligned}
& a_2=7 \times a_1 \\
&=7 \times 3=21 \\
&=3.7^{2-1} \\
& a_3=7 \times a_2 \\
&=7^2 \times a_1=49 \times 3 \\
&=147 \\
&=3.7^{3-1} \\
& a_n=7 a_{n-1} \\
&=3.7^{n-1}
\end{aligned}
$
Thus, $a_{n+1}=7 a_{n+1-1}$
$
\begin{aligned}
& =7 a_n \\
& =7 \times 3.7^{n-1} \\
& =3.7^{(m+1)-1}
\end{aligned}
$
Thus, $a_{n+1}$ is true if $a_n$ is true
Hence, by mathematical induction,
For each natural number. It is true that $a_n=3.7^{n-1}$
Question:18
Answer:
Given:
The sequence b0, b1, b2,... if defined if we let, $b_0 = 5$& for all natural numbers k. $b_k = 4 +b_{k-1}$Thus,
$b_1 = 4 + b_0\\ = 4 + 5\\ = 9 \\ =5 + 4.1$
$b_2 = 4 + b_1\\ = 4 + 9\\ = 13\\ = 5 + 4.2\\$
Now, let us consider,$b_m = 4 + b_{m-1} = 5 + 4m$ to be true.
Thus,
$b_{m+1} = 4 + b_m = 4 + 5+ 4m \\= 5 + 4(m+1)\\$
Thus, bm+1 is true if bm is true
Hence, by mathematical induction,
For each natural number n it is true that $b_n = 5 + 4n$
Question:19
Answer:
A sequence $\mathrm{d}_1, \mathrm{~d}_2, \mathrm{~d}_3 \ldots \ldots$ Is defined by letting
$
\begin{aligned}
& \mathrm{d}_1=2 \text { and } \mathrm{d}_{\mathrm{k}}=\frac{\mathrm{d}_{\mathrm{k}-1}}{\mathrm{k}} \mathrm{~d}_{\mathrm{n}}=\frac{2}{\mathrm{n}!} \text { numbers, } k \geq 2 \\
& \mathrm{~d}_k=\mathrm{d}_{k-1} / \mathrm{k}
\end{aligned}
$
$\mathrm{d}_2=\mathrm{d}_1 / 2$
$=2 / 2=2 / 2!$
$\mathrm{d}_3=\mathrm{d}_2 / 3$
$=1 / 3=2 / 3!$
Now, let us consider,
$
\mathrm{d}_m=\mathrm{d}_{m-1} / \mathrm{m}=2 / \mathrm{m}!\text { to be true. }
$
Thus,
$
\begin{aligned}
& d_{m+1}=d_{m+1-1} / m+1 \\
& =\mathrm{d}_{\mathrm{m}} / \mathrm{m}+1 \\
& =2 /(\mathrm{m}+1) \cdot \mathrm{m}! \\
& =2 /(\mathrm{m}+1)!
\end{aligned}
$
Thus, $d_{m+1}$ is true if $d_m$ is true
Hence, by mathematical induction,
For each natural no. n it is true that, $\mathrm{d}_{\mathrm{n}}=\frac{2}{\mathrm{n}!}$
Now, we'll substitute different values for $k$.
Question:20
Answer:
Given data:
$
\cos \alpha+\cos (\alpha+\beta)+\cos (\alpha+2 \beta)+\ldots+\cos (\alpha+(n-1) \beta)
$
$=\frac{\cos \left(\alpha+\left(\frac{n-1}{2}\right) \beta\right) \sin \left(\frac{n \beta}{2}\right)}{\sin \frac{\beta}{2}}$
Now, we'll substitute different values for n ,
$
\begin{aligned}
& \mathrm{n}=1 \\
& \cos (\alpha+(1-1) \beta)=\cos \alpha
\end{aligned}
$
$=\frac{\cos \left(\alpha+\left(\frac{1-1}{2}\right) \beta\right) \sin \left(\frac{\beta}{2}\right)}{\sin \left(\frac{\beta}{2}\right)}$
Thus, it's true
Now, $n=2$,
$
\cos \alpha+\cos (\alpha+(2-1) \beta)=\cos \alpha+\cos (\alpha+\beta)
$
$=\frac{\cos \left(\alpha+\left(\frac{2-1}{2}\right) \beta\right) \sin \left(\frac{2 \beta}{2}\right)}{\sin \left(\frac{\beta}{2}\right)}$
$2 \cos (\alpha+\beta / 2) \cos (\beta / 2)$
Thus, it's true.
$
\text { Now, let } \mathrm{n}=\mathrm{k} \\
{c}
\cos \alpha+\cos (\alpha+\beta)+\cos (\alpha+2 \beta)+\ldots+\cos (\alpha+(k-1) \beta)$
$=\frac{\cos \left(\alpha+\left(\frac{k-1}{2}\right) \beta\right) \sin \left(\frac{k \beta}{2}\right)}{\sin \left(\frac{\beta}{2}\right)} \text { be true } \\$
$\text { Thus, } n=k+1 \\
\cos \alpha+\cos (\alpha+\beta)+\cos (\alpha+2 \beta)+\ldots+\cos (\alpha+(k-1) \beta)+\cos (\alpha+(k+1-1) \beta)\\$
$=\frac{\cos \left(\alpha+\left(\frac{k-1}{2}\right) \beta\right) \sin \left(\frac{k \beta}{2}\right)}{\sin \left(\frac{\beta}{2}\right)}+\cos (\alpha+k \beta) \\$
$=\frac{\cos \left(\alpha+\left(\frac{k-1}{2}\right) \beta\right) \sin \left(\frac{k \beta}{2}\right)+\cos (\alpha+k \beta) \sin \left(\frac{\beta}{2}\right)}{\sin \left(\frac{\beta}{2}\right)}$
$=\frac{\sin \left(\alpha+\left(\frac{2 k-1}{2}\right) \beta\right) \sin \left(\alpha-\frac{\beta}{2}\right)+\sin (\alpha+(2 k-1) \beta / 2) \sin \left(\alpha+(2 k-1) \frac{\beta}{2}\right)}{2 \sin \left(\frac{\beta}{2}\right)}$
$ \frac{\sin \left(\alpha+\left(\frac{2 k-1}{2}\right) \beta\right) \sin \left(\alpha-\frac{\beta}{2}\right)}{2 \sin \left(\frac{\beta}{2}\right)} \\$
$ =\frac{\cos \left(\alpha+\left(\frac{k}{2}\right) \beta\right) \sin \left(\frac{(k-1) \beta}{2}\right)}{\sin \left(\frac{\beta}{2}\right)}$
Thus, $\mathrm{n}=\mathrm{k}+1$ is true
Thus, by mathematical Induction,
For each natural no. n it is true that,
$
\cos \alpha+\cos (\alpha+\beta)+\cos (\alpha+2 \beta)+\ldots+\cos (\alpha+(n-1) \beta)
$
$=\frac{\cos \left(\alpha+\left(\frac{n-1}{2}\right) \beta\right) \sin \left(\frac{n \beta}{2}\right)}{\sin \frac{\beta}{2}}$
Question:21
Answer:
Given:
$
\cos \theta \cos 2 \theta \cos 2^2 \theta \ldots \cos 2^{n-1} \theta=\frac{\sin 2^n \theta}{2^n \sin \theta}
$
Now, we'll substitute different values for n,
n = 1,
$\cos \theta=2 \sin \theta \cos \theta / 2 \sin \theta$
$=\sin 2^1 \theta / 2^1 \sin \theta$
Thus, it's true.
Now, $n=2$
$\cos \theta \cos 2 \theta=\sin 2^2 \theta / 2^2 \sin \theta$
$\begin{aligned} & =2 \sin 2 \theta \cos 2 \theta / 4 \sin \theta \\ & =4 \sin \theta \cos \theta \cos 2 \theta / 4 \sin \theta\end{aligned}$
Thus, it is true.
$
\begin{aligned}
& \quad n=3 \\
& \cos \theta \cos 2 \theta \cos 4 \theta=\sin 2^3 \theta / 2^3 \sin \theta \\
& =2 \sin 4 \theta \cos 4 \theta / 8 \sin \theta \\
& =4 \sin 2 \theta \cos 2 \theta \cos 4 \theta / 8 \sin \theta \\
& =8 \sin \theta \cos \theta \cos 2 \theta \cos 4 \theta / 8 \sin \theta
\end{aligned}
$
n = k,
$\cos \theta \cos 2 \theta \cos 2^2 \theta \ldots \cos 2^{\kappa-1} \theta=\sin 2^\kappa \theta / 2^\kappa \sin \theta$ to be true.
Thus, at $n=k+1$,
$=\cos \theta \cos 2 \theta \cos 2^2 \theta \ldots \cos 2^{k-1} \theta \cos 2^{k+1-1} \theta$
$=\sin 2^k \theta / 2^k \sin \theta \cdot \cos 2^k \theta$
$=2 \sin 2^k \theta \cos 2^k \theta / 2^{k+1} \sin \theta \\ $
$ =\sin 2^{k+1} \theta / 2^{k+1} \sin \theta$
Thus, $n=k+1$ is true
Thus, by mathematical Induction,
For each natural no. n it is true that,
$\cos \theta \cos 2 \theta \cos 2^2 \theta \ldots \cos 2^{n-1} \theta=\frac{\sin 2^n \theta}{2^n \sin \theta}$
Question:22
Answer:
Given:
P(n) : $\sin \theta+\sin 2\theta+\sin 3\theta+...+\sin n\theta\begin{aligned} &=\frac{\frac{\sin \mathrm{n} \theta}{2} \sin \frac{(\mathrm{n}+1)}{2} \theta}{\sin \frac{\theta}{2}} \end{aligned}$
Now, we’ll substitute different values for n=1,
$\\\sin \theta=[\sin (1 \theta / 2) \sin (1+1) \theta / 2] / \sin \theta / 2$
Thus, it is true.Now,
let us assume that, for n=k is true
$\sin \theta+\sin 2 \theta+\sin 3 \theta+\ldots \ldots+\sin k \theta=[\sin (k \theta / 2) \sin (k+1) \theta / 2] / \sin \theta / 2$
At n = k+1
$\sin \theta+\sin 2 \theta+\sin 3 \theta+\ldots \ldots+\sin k \theta+\sin (k+1) \theta\\$
$\\= \sin \mathrm{k} \theta / 2 \sin (\mathrm{k}+1) \theta / 2+\sin \theta / 2 \sin (\mathrm{k}+1) \Theta] / \sin \theta / 2$ $\\ =[\cos (\theta / 2)-\cos (2 k+1) \theta / 2+\cos (2 k+1) \theta / 2-\cos (2 k+3) \theta / 2] / 2 \sin \theta / 2$ $\\ =[\cos (\theta / 2)-\cos (2 k+3) \Theta / 2] / 2 \sin \theta / 2$ $\\ =[\sin \mathrm{k} \theta / 2 \sin (\mathrm{k}+1) \ominus / 2+\sin \theta / 2 \sin (\mathrm{k}+1) \Theta] / \sin \theta / 2$
$\\ =\sin \mathrm{k} \theta / 2 \sin (\mathrm{k}+1) \theta / 2+\sin \theta / 2 \sin (\mathrm{k}+1) \Theta] / \sin \theta / 2$ $\\ =[\cos (\theta / 2)-\cos (2 k+1) \theta / 2+\cos (2 k+1) \theta / 2-\cos (2 k+3) \theta / 2] / 2 \sin \theta / 2$ $\\ =[\cos (\theta / 2)-\cos (2 k+3) \Theta / 2] / 2 \sin \theta / 2$ $\\ =[\sin \mathrm{k} \theta / 2 \sin (\mathrm{k}+1) \ominus / 2+\sin \theta / 2 \sin (\mathrm{k}+1) \Theta] / \sin \theta / 2$
Thus, n=k+1 is true
Thus, by mathematical Induction,
For each natural no. n it is true that,
$\sin \theta+\sin 2\theta+\sin 3\theta+...+\sin n\theta\begin{aligned} &=\frac{\frac{\sin \mathrm{n} \theta}{2} \sin \frac{(\mathrm{n}+1)}{2} \theta}{\sin \frac{\theta}{2}} \end{aligned}$
Question:23
Show that $\frac{n^{5}}{5}+\frac{n^{3}}{3}+\frac{7n}{15}$ is a natural number for all n $\epsilon$N.
Answer:
Given data:
P(n) = $\frac{n^{5}}{5}+\frac{n^{3}}{3}+\frac{7n}{15}$
Now, we’ll substitute different values for n,
$\text { At } n=1 $
$\\ P(1)=\frac{1^{5}}{5}+\frac{1^{8}}{3}+\frac{7(1)}{15} \\ $
$=\frac{1}{5}+\frac{1}{3}+\frac{7}{15}$
$\\=(3+5+7 )/ 15\\ =1 \: \: is\: \: true.\\\\\ $
$At \: \:$
$n=2$
$ P(2)=\frac{2^{5}}{5} +\frac{2^{3}}{3}+\frac{7(2)}{15} \\\\$
$ =\frac{32}{5}+\frac{8}{3}+\frac{14}{15} \\\\ =(96+40+14 )/ 15 =10 \: \: is\: \: true$
At n=3
$\\P(3)=\frac{3^{5}}{5} +\frac{3^{3}}{3}+\frac{7(3)}{15} \\\\ $
$=\frac{243}{5}+\frac{27}{3}+\frac{21}{15} \\\\ =(729+135+21 )/ 15 \\\\$
$=59$ is true.
Let us consider,
$P(k)=\frac{k^{5}}{5}+\frac{k^{2}}{3}+\frac{7(k)}{15}$$\: \: to\: \: be\: \: true.$
Thus, at n=k+1
$ P(k+1)=\frac{(k+1)^5}{5}+\frac{(k+1)^3}{3}+\frac{7(k+1)}{15} \\ $
$ =\frac{k^5+5 k^4+10 k^2+10 k^2+5 k+1}{5}+\frac{k^3+3 k^2+3 k+1}{3}+\frac{7(k+1)}{15} \\ $
$=\frac{k^5}{5}+\frac{k^3}{3}+\frac{7(k)}{15}+k^4+2 k^3+3 k^2+2 k+1, \text { is a natural number }$
Thus, n = k + 1 is true
Thus, by mathematical Induction,
For each natural number. It is true that,
P(n) = $\frac{n^{5}}{5}+\frac{n^{3}}{3}+\frac{7n}{15}$
Question:24
Answer:
Given:
P(n) = $\frac{1}{n+1}+\frac{1}{n+2}+\ldots .+\frac{1}{2 n}>\frac{13}{24}$
Now, we’ll substitute different values for n,
$ \text { At } n=2 \\$
$ P(2)=1 / 2 +1 / 4 \\ =3 / 4>13 / 24 \\$
$ It\: \: is\: \: true.\\ $
$At \: \: $n=3$ \\ $
$P(3)=1 / 2+1 / 4+1 / 6 \\ =11 / 12>13 / 24 \\$.
It is true.
Now, let us consider,
$P(k)=\frac{1}{k+1}+\frac{1}{k+2}+\ldots \ldots+\frac{1}{2 k}>\frac{13}{24}$ To be true.
Now, at $n=k+1$,
$
P(k+1)=\frac{1}{(k+1)+1}+\frac{1}{(k+1)+2}+\ldots . .+\frac{1}{2(k+1)}
$
$\begin{array}{l} =\frac{1}{k+2}+\frac{1}{k+3} \ldots \ldots+\frac{1}{2 k+2} \\\\ =\frac{1}{k+1}+\frac{1}{k+2}+\frac{1}{k+3} \ldots \ldots+\frac{1}{2 k}+\frac{1}{2 k+1}+\frac{1}{2 k+2}-\frac{1}{k+1} \\\\ =\frac{1}{k+1}+\frac{1}{k+2}+\frac{1}{k+3} \ldots \ldots+\frac{1}{2 k}+\frac{2k+2+2k+1-4k-2}{2(2 k+1)(k+1)} \\\\ =\frac{1}{k+1}+\frac{1}{k+2}+\frac{1}{k+3} \ldots \ldots+\frac{1}{2 k}+\frac{1}{2(2 k+1)(k+1)}>\frac{13}{24} \end{array}$
Thus, n = k + 1 is true
Thus, by mathematical Induction,
For each natural no. n > 1 it is true that,
P(n) = $\frac{1}{n+1}+\frac{1}{n+2}+\ldots .+\frac{1}{2 n}>\frac{13}{24}$
Question:25
Answer:
To Prove:
$2^n$ is the no. of subsets of a set containing n distinct elements.
Now, there is only one subset since for a null set, there is only one element φ
Now, we’ll substitute different values for n,
At n = 0,
No. of subsets = 1 = $2^0$
If we consider a set that has 1 element, then there will be 2 subsets that have 1 & φ
Thus, at n = 1, it is true
now at n = 2
No. of subsets = 2 = $2^2$
Now, let us consider that there is a set $S_k$ that has k elements.
At n = k,
No. of subsets = $2^k$ is true.
Now, for set $S_{k+1}$, having k+1 elements,
Compared to the set $S_{k}$, $S_{k+1}$ has an extra element and thus will form an extra collection of subsets when combined with the existing $2^k$ subsets & an extra $2^k$ subset will be formed.
Thus, at n = k+1,
No. of subsets = $2^k$ + $2^k$
= 2.$2^k$
= $2^{k+1}$
Thus, by mathematical Induction,
$2^n$ is the no. of subsets of a set containing n distinct elements.
Question:26
If $10^n + 3.4^{n+2} + k$ is divisible by 9 for all n? N, then the least positive integral value of k is
A. 5
B. 3
C. 7
D. 1
Answer:
The answer is option (A)
Given:
P(n) = $10^n + 3.4^{n+2} + k$ is divisible by 9
Now, we’ll substitute different values for n,
At n = 1,
$P(1) = 10^1 + 3.4^3+ k$
$= 202 + k$
Now, we know that the sum of the digits of this no should be 9 or multiples of 9, for it to be divisible by 9.
Thus $2 + 2 + k = 9$
Thus,$k = 5.$
Question:27
For all n? N, $3.5^{2n + 1} + 2^{3n + 1}$ is divisible by
A. 19
B. 17
C. 23
D. 25
Answer:
The answer is option (B)
Given:
P(n) = $3.5^{2n + 1} + 2^{3n + 1}$
Now, we’ll substitute different values for n,
At n = 1,
P(1) = $3.5^{2+1} + 2^{3+1}$
$\\ = 375 + 16\\ = 391 \\=17\times 23$
At n = 2,
$P(2) = 3.5^{4+1} + 2^{6+1}$
$\\= 9375 + 256 \\ = 9503 \\ = 17 \times 559$
Therefore, it is divisible by 17.
Question:28
If $x^n-1$ is divisible by$x - k$ , then the least positive integral value of k is
A. 1
B. 2
C. 3
D. 4
Answer:
The answer is option (A)
Given:
P(n) = $x^n-1$ is divisible by$x - k$
Now, we’ll substitute different values for n,
At n = 1,
$P(1) = x -1$
At n = 2,
$P(2) = x^2 -1$
$= (x -1) (x + 1)$
At n = 3,
$P(3) = x^3 -1$
$= (x -1) (x^2 + xy + 1)$
At n = 4,
$\\P(4) = x^4 -1\\ = (x^2 -1) (x^2 + 1) \\ = (x -1) (x + 1) (x^2 + 1)$
Therefore, ‘1’ is the least possible integral value of k.
Question:29
Answer:
$n\geq4$
Given:
$P(n) : 2n < n!, n \in N$
Now, we’ll substitute different values for n,
At n = 1,
$P(1) = 2 \times 1 < 1!$,
$= 2 < 1$, viz. not true.
At n = 2,
$P(2) = 2 \times 2 < 2!$
$= 4 < 2$, viz. not true.
At n = 3,
$P(3) = 2\times 3 < 3!$
$= 6 < 6$, viz. not true
At n = 4,
$P(4) = 2 \times 4 < 4!$
$= 8 < 24$, viz. not true
At n = 5,
$P(5) = 2 \times 5 < 5!$
$= 10 < 120$, viz. true
Therefore, $n\geq4$
Question:30
Answer:
True.
Explanation: It is the Principle of Mathematical Induction.
These solutions will ultimately help students develop skills regarding inductive and deductive reasoning used in almost every other field, ranging from science to many other branches.
Here are the subject-wise links for the NCERT solutions of class 11:
Given below are the subject-wise NCERT Notes of class 11 :
Here are some useful links for NCERT books and the NCERT syllabus for class 11:
Given below are the subject-wise exemplar solutions of class 11 NCERT:
To prove that a given statement is true for all natural numbers, the Principle of Mathematical Induction is applied. It is based on two essential steps:
To answer questions on chapter 4 (Mathematical Induction ) of the NCERT Exemplar, first, carefully read the question to find out what's on the other side. Get down to business by proving the basic case, where you check that the statement is correct for a small amount.
n, often
n=1. Next, assume that the statement is true for some
n=k, that's the inductive theory. In the Inductive move, use the presumption to show that the assertion includes.
n=k+2. If you succeed in proving it, you have revealed that this observation applies to all inherent numbers. To solve these problems, you need practice, as you need to carefully work by taking all the steps to ensure that the initiation method is properly followed.
This chapter includes solving a result using the principle of mathematical induction. It uses the basic situation to derive the result and then applies it for higher values.
The process of Mathematical Induction follows a structured, step-by-step approach. First, you begin by proving the Base Case, where you demonstrate that the statement is true for the smallest value, typically n=1. Next, in the Inductive Hypothesis, you assume that the statement is true for some arbitrary value n=k. This assumption is then used to prove the Inductive Step, where you show that the statement holds for n=k+1. Once you have successfully proven both the base case and the inductive step, you can conclude, by the principle of induction, that the statement is true for all natural numbers. The step-by-step nature of this process ensures that the proof is rigorous and logical.
The Principle of Mathematical Induction is a powerful tool used to prove statements that are true for all natural numbers. It is commonly used to prove formulas involving sums, products, or other expressions that hold for all values of nnn. For example, induction is used to prove statements such as the sum of the first nnn natural numbers or the formula for the sum of squares. It is also applied to prove divisibility rules, inequalities, and properties of sequences. By breaking the proof into two manageable steps—the base case and the inductive step—induction provides a structured way to establish the truth of statements across an infinite set of natural numbers. This method is widely applicable in various areas of mathematics, making it an essential tool in mathematical reasoning and proof construction.
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