NCERT Solutions for Exercise 8.1 Class 11 Maths Chapter 8

NCERT Solutions for Exercise 8.1 Class 11 Maths Chapter 8 - Binomial Theorem

Edited By Ravindra Pindel | Updated on Jul 13, 2022 03:40 PM IST

In the earlier classes, you have already learned about finding the squares and cubes of (a+b) and (a-b) using repeated multiplications but if the power is higher it would be very difficult to find the value of (a+b)n using repeated multiplications. The binomial theorem is the solution to find the value of (a+b)ⁿ. In the NCERT solutions for Class 11 Maths chapter 8 exercise 8.1 where you will learn about the binomial expansion of (a+b)ⁿ.

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Binomial Theorem Class 11 Chapter 8 Exercise: 8.1
Question:1 Expand the expression.
More About NCERT Solutions for Class 11 Maths Chapter 8 Exercise 8.1:-
NCERT Solutions of Class 11 Subject Wise
Subject Wise NCERT Exampler Solutions

You will learn about the pascal triangle and binomial expansion for the positive integers in exercise 8.1 Class 11 Maths. In the previous chapter, you have learned about factorial expansion which will be useful in the binomial expansion. There are some observation points given before the Class 11 Maths chapter 8 exercise 8.1. You must go through these points in order to get conceptual clarity. If you are looking for NCERT Solutions, you can go through the given link where you will get NCERT solutions for Classes 6 to 12.

Also, see

Binomial Theorem Class 11 Chapter 8 Exercise: 8.1

Question:1 Expand the expression. $(1-2x)^5$

Answer:

Given,

The Expression:

$(1-2x)^5$

the expansion of this Expression is,

$(1-2x)^5 =$

$\\^5C_0(1)^5-^5C_1(1)^4(2x)+^5C_2(1)^3(2x)^2-^5C_3(1)^2(2x)^3+^5C_4(1)^1(2x)^4-^5C_5(2x)^5$

$1-5(2x)+10(4x^2)-10(8x^3)+5(16x^4)-(32x^5)$

$1-10x+40x^2-80x^3+80x^4-32x^5$

Question:2 Expand the expression. $\left(\frac{2}{x} - \frac{x}{2} \right )^5$

Answer:

Given,

The Expression:

$\left(\frac{2}{x} - \frac{x}{2} \right )^5$

the expansion of this Expression is,

$\left(\frac{2}{x} - \frac{x}{2} \right )^5\Rightarrow$

$\\^5C_0\left(\frac{2}{x}\right)^5-^5C_1\left(\frac{2}{x}\right)^4\left(\frac{x}{2}\right)+^5C_2\left(\frac{2}{x}\right)^3\left(\frac{x}{2}\right)^2$ $-^5C_3\left(\frac{2}{x}\right)^2\left(\frac{x}{2}\right)^3+^5C_4\left(\frac{2}{x}\right)^1\left(\frac{x}{2}\right)^4-^5C_5\left(\frac{x}{2}\right)^5$

$\Rightarrow \frac{32}{x}-5\left ( \frac{16}{x^4} \right )\left ( \frac{x}{2} \right )+10\left ( \frac{8}{x^3} \right )\left ( \frac{x^2}{4} \right )-10\left ( \frac{4}{x^2} \right )\left ( \frac{x^2}{8} \right )$ $+5\left ( \frac{2}{x} \right )\left ( \frac{x^4}{16} \right )-\frac{x^5}{32}$

$\Rightarrow \frac{32}{x^5}-\frac{40}{x^3}+\frac{20}{x}-5x+\frac{5x^2}{8}-\frac{x^3}{32}$

Question:3 Expand the expression. $(2x-3)^6$

Answer:

Given,

The Expression:

$(2x-3)^6$

the expansion of this Expression is,

$(2x-3)^6=$

$\Rightarrow$ $\\^6C_0(2x)^6-^6C_1(2x)^5(3)+^6C_2(2x)^4(3)^2-^6C_3(2x)^3(3)^3+$ $^6C_4(2x)^2(3)^4-^6C_5(2x)(3)^5+^6C_6(3)^6$

$\Rightarrow$ $64x^6-6(32x^5)(3)+15(16x^4)(9)-20(8x^3)(27)+15(4x^2)(81)-6(2x)(243)$ $+729$

$\Rightarrow$ $64x^6-576x^5+2160x^4-4320x^3+4860x^2-2916x+729$

Question:4 Expand the expression. $\left(\frac{x}{3} + \frac{1}{x} \right )^5$

Answer:

Given,

The Expression:

$\left(\frac{x}{3} + \frac{1}{x} \right )^5$

the expansion of this Expression is,

$\left(\frac{x}{3} + \frac{1}{x} \right )^5\Rightarrow$

$\Rightarrow ^5C_0\left(\frac{x}{3}\right)^5+^5C_1\left(\frac{x}{3}\right)^4\left(\frac{1}{x}\right)+^5C_2\left(\frac{x}{3}\right)^3\left(\frac{1}{x}\right)^2$ $+^5C_3\left(\frac{x}{3}\right)^2\left(\frac{1}{x}\right)^3+^5C_4\left(\frac{x}{3}\right)^1\left(\frac{1}{x}\right)^4+^5C_5\left(\frac{1}{x}\right)^5$

$\Rightarrow \frac{x^5}{243} +5\left ( \frac{x^4}{81} \right )\left ( \frac{1}{x} \right )+10\left ( \frac{x^3}{27} \right )\left ( \frac{1}{x^2} \right )+10\left ( \frac{x^2}{9} \right )\left ( \frac{1}{x^3} \right )$ $+5\left ( \frac{x}{3} \right )\left ( \frac{1}{x^4} \right )+\frac{1}{x^5}$

$\Rightarrow \frac{x^5}{243}+\frac{5x^3}{81}+\frac{10x}{27}+\frac{10}{9x}+\frac{5}{3x^2}+\frac{1}{x^5}$

Question:5 Expand the expression. $\left(x + \frac{1}{x} \right )^6$

Answer:

Given,

The Expression:

$\left(x + \frac{1}{x} \right )^6$

the expansion of this Expression is,

$\left(x + \frac{1}{x} \right )^6$

$\Rightarrow ^6C_0(x)^6+^6C_1(x)^5\left ( \frac{1}{x} \right )+^6C_2(x)^4\left ( \frac{1}{x} \right )^2+^6C_3(x)^3\left ( \frac{1}{x} \right )^3+$ $^6C_4(x)^2\left ( \frac{1}{x} \right )^4+^6C_5(x)\left ( \frac{1}{x} \right )^5+^6C_6\left ( \frac{1}{x} \right )^6$

$\Rightarrow x^6+6(x^5)\left ( \frac{1}{x} \right )+15(x^4)\left ( \frac{1}{x^2} \right )+20(8x^3)\left ( \frac{1}{x^3} \right )$ $+15(x^2)\left ( \frac{1}{x^4} \right )+6(x)\left ( \frac{1}{x^5} \right )+\frac{1}{x^6}$

$\Rightarrow x^6+6x^4+15x^2+20+\frac{15}{x^2}+\frac{6}{x^4}+\frac{1}{x^6}$

Question:6 Using binomial theorem, evaluate the following: $(96)^3$

Answer:

As 96 can be written as (100-4);

$\\\Rightarrow (96)^3\\=(100-4)^3\\=^3C_0(100)^3-^3C_1(100)^2(4)+^3C_2(100)(4)^2-^3C_3(4)^3$

$\\=(100)^3-3(100)^2(4)+3(100)(4)^2-(4)^3$

$\\=1000000-120000+4800-64$

$\\=884736$

Question:7 Using binomial theorem, evaluate the following: $(102)^5$

Answer:

As we can write 102 in the form 100+2

$\Rightarrow (102)^5$

$=(100+2)^5$

$\\=^5C_0(100)^5+^5C_1(100)^4(2)+^5C_2(100)^3(2)^2$ $+^5C_3(100)^2(2)^3+^5C_4(100)^1(2)^4+^5C_5(2)^5$

$=10000000000+1000000000+40000000+800000+8000+32$

$=11040808032$

Question:8 Using binomial theorem, evaluate the following:

$(101)^4$

Answer:

As we can write 101 in the form 100+1

$\Rightarrow (101)^4$

$=(100+1)^4$

$\\=^4C_0(100)^4+^4C_1(100)^3(1)+^4C_2(100)^2(1)^2$ $+^4C_3(100)^1(1)^3+^4C_4(1)^4$

$=100000000+4000000+60000+400+1$

$=104060401$

Question:9 Using binomial theorem, evaluate the following: $(99)^5$

Answer:

As we can write 99 in the form 100-1

$\Rightarrow (99)^5$

$=(100-1)^5$

$\\=^5C_0(100)^5-^5C_1(100)^4(1)+^5C_2(100)^3(1)^2$ $-^5C_3(100)^2(1)^3+^5C_4(100)^1(1)^4-^5C_5(1)^5$

$=10000000000-500000000+10000000-100000+500-1$

$=9509900499$

Question:10 Using Binomial Theorem, indicate which number is larger (1.1)¹⁰⁰⁰⁰ or 1000.

Answer:

AS we can write 1.1 as 1 + 0.1,

$(1.1)^{10000}=(1+0.1)^{10000}$

$=^{10000}C_0+^{10000}C_1(1.1)+Other \:positive\:terms$

$=1+10000\times1.1+ \:Other\:positive\:term$

$>1000$

Hence,

$(1.1)^{10000}>1000$

Question:11 Find $(a + b)^4 - (a-b)^4$ . Hence, evaluate $(\sqrt{3} + \sqrt2)^4 - (\sqrt3-\sqrt2)^4$ .

Answer:

Using Binomial Theorem, the expressions $(a+b)^4$ and $(a-b)^4$ can be expressed as

$(a+b)^4=^4C_0a^4+^4C_1a^3b+^4C_2a^2b^2+^4C_3ab^3+^4C_4b^4$

$(a-b)^4=^4C_0a^4-^4C_1a^3b+^4C_2a^2b^2-^4C_3ab^3+^4C_4b^4$

From Here,

$(a+b)^4-(a-b)^4 = ^4C_0a^4+^4C_1a^3b+^4C_2a^2b^2+^4C_3ab^3+^4C_4b^4$ $-^4C_0a^4+^4C_1a^3b-^4C_2a^2b^2+^4C_3ab^3-^4C_4b^4$

$(a+b)^4-(a-b)^4 = 2\times( ^4C_1a^3b+^4C_3ab^3)$

$(a+b)^4-(a-b)^4 = 8ab(a^2+b^2)$

Now, Using this, we get

$(\sqrt{3} + \sqrt2)^4 - (\sqrt3-\sqrt2)^4=8(\sqrt{3})(\sqrt{2})(3+2)=8\times\sqrt{6}\times5=40\sqrt{6}$

Question:12 Find $(x+1)^6 + (x-1)^6$ . Hence or otherwise evaluate $(\sqrt2+1)^6 + (\sqrt2-1)^6$ .

Answer:

Using Binomial Theorem, the expressions $(x+1)^4$ and $(x-1)^4$ can be expressed as ,

$(x+1)^6=^6C_0x^6+^6C_1x^51+^6C_2x^41^2+^4C_3x^31^3+^6C_4x^21^4+^6C_5x1^5+^6C_61^6$

$(x-1)^6=^6C_0x^6-^6C_1x^51+^6C_2x^41^2-^4C_3x^31^3+^6C_4x^21^4-^6C_5x1^5+^6C_61^6$

From Here,

$\\(x+1)^6-(x-1)^6=^6C_0x^6+^6C_1x^51+^6C_2x^41^2+^4C_3x^31^3+$ $^6C_4x^21^4+^6C_5x1^5+^6C_61^6$ $\:\:\:\:\;\:\:\;\:\:\:\ +^6C_0x^6-^6C_1x^51+^6C_2x^41^2-^4C_3x^31^3+^6C_4x^21^4-^6C_5x1^5+^6C_61^6$

$(x+1)^6+(x-1)^6=2(^6C_0x^6+^6C_2x^41^2+^6C_4x^21^4+^6C_61^6)$

$(x+1)^6+(x-1)^6=2(x^6+15x^4+15x^2+1)$

Now, Using this, we get

$(\sqrt2+1)^6 + (\sqrt2-1)^6=2((\sqrt{2})^6+15(\sqrt{2})^4+15(\sqrt{2})^2+1)$

$(\sqrt2+1)^6 + (\sqrt2-1)^6=2(8+60+30+1)=2(99)=198$

Question:13 Show that $9^{n+1} - 8n - 9$ is divisible by 64, whenever n is a positive integer.

Answer:

If we want to prove that $9^{n+1} - 8n - 9$ is divisible by 64, then we have to prove that $9^{n+1} - 8n - 9=64k$

As we know, from binomial theorem,

$(1+x)^m=^mC_0+^mC_1x+^mC_2x^2+^mC_3x^3+....^mC_mx^m$

Here putting x = 8 and replacing m by n+1, we get,

$9^{n+1}=^{n+1}C_0+\:^{n+1}C_18+^{n+1}C_28^2+.......+^{n+1}C_{n+1}8^{n+1}$

$9^{n+1}=1+8(n+1)+8^2(^{n+1}C_2+\:^{n+1}C_38+^{n+1}C_48^2+.......+^{n+1}C_{n+1}8^{n-1})$

$9^{n+1}=1+8n+8+64(k)$

Now, Using This,

$9^{n+1} - 8n - 9=9+8n+64k-9-8n=64k$

Hence

$9^{n+1} - 8n - 9$ is divisible by 64.

Question:14 Prove that $\sum_{r = 0}^n3^r \ ^nC_r = 4^n$

Answer:

As we know from Binomial Theorem,

$\sum_{r = 0}^na^r \ ^nC_r = (1+a)^n$

Here putting a = 3, we get,

$\sum_{r = 0}^n3^r \ ^nC_r = (1+3)^n$

$\sum_{r = 0}^n3^r \ ^nC_r = 4^n$

Hence Proved.

NCERT Solutions of Class 11 Subject Wise

Subject Wise NCERT Exampler Solutions

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Frequently Asked Questions (FAQs)

1. What is the weightage of co-ordinate geometry in the CBSE Class 11 Maths ?

Co-ordinate geometry has 10 marks weightage in the CBSE Class 11 Maths exam.

2. Where can I get NCERT exemplar solutions for Class 11 free?

Here you will get NCERT Exemplar Solutions for Class 11 Free.

3. Are these NCERT exercise solutions good for revision ?

Yes, NCERT exercise solutions are very helpful for the revision of the important concepts before the exam.

4. Who prepared the NCERT exercise wise solutions ?

NCERT exercise-wise solutions are prepared by subject matter experts who are experienced in this field.

5. What are the important topics covered in the Class 11 Maths exercise 8.1 ?

Binomial expansion for positive integers and pascal triangle are covered in the Class 11 Maths exercise 8.1.

6. Does NCERT solutions for Class 11 Maths contains diagrams ?

Yes, NCERT solutions for Class 11 Maths contain the diagrams and charts where ever it is necessary.

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NCERT Exemplar Class 11 Maths Solutions Chapter 7 Permutations and Combinations

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