# NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem

NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem: You have studied the expansion of expressions like (a-b)2 and (a-b)in the previous classes. So you can calculate numbers like (96)3. If the power is high, it will be difficult to use normal multiplication. How will you process in such cases? In the NCERT solutions for class 11 maths chapter 8 binomial theorem, you will get the answer to the above question. In this chapter, you will study the expansion of (a+b)n, the general terms of the expansion, the middle term of the expansion, and the pascal triangle. In solutions of NCERT for class 11 chapter 8 binomial theorem, you will get questions related to these topics. This chapter covers the binomial theorem for positive integral indices only. The concepts of a binomial theorem are not only useful in solving problems of mathematics, but in various fields of science too. In this chapter, there are  26 problems in 2 exercises. All these questions are prepared in NCERT solutions for class 11 maths chapter 8 binomial theorem in a detailed manner. It will be very easy for you to understand the concepts. Check all NCERT solutions from class 6 to 12 to learn science and maths. There are 2 exercise and a miscellaneous exercise in this chapter which are explained below.

Exercise:8.1

Exercise:8.2

Miscellaneous Exercise

## The main content headings of NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem are listed below:

8.1 Introduction

8.2 Binomial Theorem for Positive Integral Indices

8.3 General and Middle Terms

The concepts of NCERT Class 11 Maths Chapter 8 Binomial Theorem can be used to find the approximate value of the power of a small number. For example, find the approximate value of 0.996 using the first three terms of expansion? This can be solved by rewriting 0.996 as (1-0.01)6 and expanding using the Binomial Theorem.

## NCERT solutions for class 11 maths chapter 8 binomial theorem-Exercise: 8.1

Given,

The Expression:

$(1-2x)^5$

the expansion of this Expression is,

$(1-2x)^5 =$

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

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

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

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}$

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

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$

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}$

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

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}$

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

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

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

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

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

$\\=884736$

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$

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

$=11040808032$

$(101)^4$

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$

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

$=104060401$

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$

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

$=9509900499$

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,

$(1.1)^{10000}>1000$

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

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

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

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

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

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.

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.

## Question:1 Find the coefficient of

$x^5$ in $(x + 3)^8$

As we know that the $(r+1)^{th}$ term  $T_{r+1}$ in the binomial expansion of  $(a+b)^n$  is given by

$T_{r+1}=^nC_ra^{n-r}b^r$

Now let's assume $x^5$ happens in the $(r+1)^{th}$ term of the binomial expansion of $(x + 3)^8$

So,

$T_{r+1}=^8C_rx^{8-r}3^r$

On comparing the indices of x we get,

$r=3$

Hence the coefficient of the  $x^5$ in $(x + 3)^8$ is

$^8C_3\times3^3=\frac{8!}{5!3!}\times 9=\frac{8\times7\times6}{3\times2}\times9=1512$

As we know that the $(r+1)^{th}$ term  $T_{r+1}$ in the binomial expansion of  $(a+b)^n$  is given by

$T_{r+1}=^nC_ra^{n-r}b^r$

Now let's assume $a^5b^7$ happens in the $(r+1)^{th}$ term of the binomial expansion of $(a- 2b)^{12}$

So,

$T_{r+1}=^{12}C_rx^{12-r}(-2b)^r$

On comparing the indices of x we get,

$r=7$

Hence the coefficient of the  $a^5b^7$  in $(a- 2b)^{12}$ is

$\\ \Rightarrow ^{12}C_7\times(-2)^7=\frac{12!}{5!7!}\times (-128)\\=\frac{12\times11\times10\times 9\times8}{5\times4\times3\times2}\times(-128) \\=-(729)(128) \\=-101376$

$(x^2 - y)^6$

As we know that the general  $(r+1)^{th}$ term  $T_{r+1}$ in the binomial expansion of  $(a+b)^n$  is given by

$T_{r+1}=^nC_ra^{n-r}b^r$

So the general term of the expansion of $(x^2 - y)^6$  :

$T_{r+1}=^6C_r(x^2)^{6-r}(-y)^r=(-1)^r\times^6C_rx^{12-2r}y^r$.

$(x^2 - xy)^{12}, \ x\neq 0$

As we know that the general  $(r+1)^{th}$ term  $T_{r+1}$ in the binomial expansion of  $(a+b)^n$  is given by

$T_{r+1}=^nC_ra^{n-r}b^r$

So the general term of the expansion of $(x^2 - xy)^{12},$ is

$\\\Rightarrow T_{r+1}\\=^{12}C_r(x^2)^{12-r}(-xy)^r\\=(-1)^r\times^{12}C_rx^{24-2r+r}y^r\\=(-1)^r\times^{12}C_rx^{24-r}y^r$.

As we know that the general  $(r+1)^{th}$ term  $T_{r+1}$ in the binomial expansion of  $(a+b)^n$  is given by

$T_{r+1}=^nC_ra^{n-r}b^r$

So the $4^{th}$ term of the expansion of $(x-2y)^{12}$ is

$\\\Rightarrow T_4= T_{3+1}\\=^{12}C_3(x)^{12-3}(-2y)^3\\=(-2)^3\times^{12}C_3x^{9}y^3\\=-8\times\frac{12!}{3!9!}x^{9}y^3$

$\\=-8\times \frac{12\times11\times10}{3\times2}\times x^9y^3\\=-8\times220\times x^9y^3 \\=-1760x^9y^3$.

As we know that the general  $(r+1)^{th}$ term  $T_{r+1}$ in the binomial expansion of  $(a+b)^n$  is given by

$T_{r+1}=^nC_ra^{n-r}b^r$

So the $13^{th}$ term of the expansion of     $\left(9x - \frac{1}{3\sqrt x} \right )^{18}$   is

$\\\Rightarrow T_{13}= T_{12+1}\\=^{18}C_{12}(9x)^{18-12}\left(\frac{1}{3\sqrt{x}}\right)^{12}\\ \\ \\=\frac{18!}{12!6!}\times9^{6}\left ( \frac{1}{3} \right )^{12}\times x^{6-6}$

$\\=\frac{18\times17\times16\times15\times14\times13}{6\times5\times4\times3\times2}\times9^{6}\left ( \frac{1}{3^{12}} \right )$

$=18564$

As we know that the middle  terms in the expansion of  $(a+b)^n$ when n is odd are,

$\left ( \frac{n+1}{2} \right )^{th}\:term\: \:and\:\:\left ( \frac{n+1}{2}+1 \right )^{th}\:term$

Hence the middle term of the expansion    $\left(3 - \frac{x^3}{6} \right )^7$   are

$\left ( \frac{7+1}{2} \right )^{th}\:term\: \:and\:\:\left ( \frac{7+1}{2}+1 \right )^{th}\:term$

Which are $4^{th}\:term\:and\:\:5^{th} \:term$

Now,

As we know that the general  $(r+1)^{th}$ term  $T_{r+1}$ in the binomial expansion of  $(a+b)^n$  is given by

$T_{r+1}=^nC_ra^{n-r}b^r$

So the $4^{th}$ term of the expansion of   $\left(3 - \frac{x^3}{6} \right )^7$ is

$\\\Rightarrow T_4= T_{3+1}\\=^{7}C_3(3)^{7-3}\left (- \frac{x^3}{6} \right )^3\\=\left ( -\frac{1}{6} \right )^3\times 3^4\times^{7}C_3\times x^{9}\\=\left ( -\frac{1}{6} \right )^3\times 3^4\times\frac{7!}{3!4!}\times x^{9} \\=-\frac{3\times3\times3\times3}{6\times6\times6}\times\frac{7\times 6\times5}{3\times2}\times x^9$

$=-\frac{105}{8}x^9$

And the $5^{th}$ Term of the expansion of   $\left(3 - \frac{x^3}{6} \right )^7$  is,

$\\\Rightarrow T_5= T_{4+1}\\=^{7}C_4(3)^{7-4}\left (- \frac{x^3}{6} \right )^4\\=\left ( -\frac{1}{6} \right )^4\times 3^3\times^{7}C_4\times x^{12}\\=\left ( -\frac{1}{6} \right )^4\times 3^3\times\frac{7!}{3!4!}\times x^{12} \\=\frac{3\times3\times3}{6\times6\times6\times6}\times\frac{7\times 6\times5}{3\times2}\times x^{12}$

$=\frac{35}{48}x^{12}$

Hence the middle terms of the expansion of given expression are

$-\frac{105}{8}x^9\:and\:\frac{35}{48}x^{12}.$

As we know that the middle term in the expansion of  $(a+b)^n$ when n is even is,

$\left ( \frac{n}{2}+1 \right )^{th}\:term\:$,

Hence the middle term of the expansion     $\left(\frac{x}{3} + 9y \right )^{10}$   is,

$\left ( \frac{10}{2}+1 \right )^{th}\:term\:$

Which is $6^{th}\:term$

Now,

As we know that the general  $(r+1)^{th}$ term  $T_{r+1}$ in the binomial expansion of  $(a+b)^n$  is given by

$T_{r+1}=^nC_ra^{n-r}b^r$

So the $6^{th}$ term of the expansion of   $\left(\frac{x}{3} + 9y \right )^{10}$ is

$\\\Rightarrow T_6= T_{5+1}\\=^{10}C_5\left ( \frac{x}{3} \right )^{10-5}\left ( 9y \right )^5\\$

$=\left ( \frac{1}{3} \right )^5\times9^5\times^{10}C_5\times x^5y^5$

$=\left ( \frac{1}{3} \right )^5\times9^5\times\left ( \frac{10!}{5!5!} \right )\times x^5y^5$

$=\left ( \frac{1}{3^5} \right )\times9^5\times\left ( \frac{10\times9\times8\times7\times6}{5\times4\times3\times2} \right )\times x^5y^5$

$=61236x^5y^5$

Hence the middle term of the expansion of  $\left(\frac{x}{3} + 9y \right )^{10}$ is nbsp; $61236x^5y^5$.

As we know that the general  $(r+1)^{th}$ term  $T_{r+1}$ in the binomial expansion of  $(a+b)^n$  is given by

$T_{r+1}=^nC_ra^{n-r}b^r$

So, the general $(r+1)^{th}$ term  $T_{r+1}$ in the binomial expansion of  $(1 + a)^{m+n}$  is given by

$T_{r+1}=^{m+n}C_r1^{m+n-r}a^r=^{m+n}C_ra^r$

Now, as we can see $a^m$ will come when $r=m$ and $a^n$ will come when $r=n$

So,

Coefficient of $a^m$ :

$K_{a^m}=^{m+n}C_m=\frac{(m+n)!}{m!n!}$

CoeficientCoefficient of $a^n$ :

$K_{a^n}=^{m+n}C_n=\frac{(m+n)!}{m!n!}$

As we can see $K_{a^m}=K_{a^n}$.

Hence it is proved that the coefficients of $a^m$ and $a^n$ are equal.

As we know that the general  $(r+1)^{th}$ term  $T_{r+1}$ in the binomial expansion of  $(a+b)^n$  is given by

$T_{r+1}=^nC_ra^{n-r}b^r$

So,

$(r+1)^{th}$ Term in  the expansion of  $(x+1)^{n}$:

$T_{r+1}=^nC_rx^{n-r}1^r=^nC_rx^{n-r}$

$r^{th}$ Term in  the expansion of  $(x+1)^{n}$:

$T_{r}=^nC_{r-1}x^{n-r+1}1^{r-1}=^nC_{r-1}x^{n-r+1}$

$(r-1)^{th}$ Term in  the expansion of  $(x+1)^{n}$:

$T_{r-1}=^nC_{r-2}x^{n-r+2}1^{r-2}=^nC_{r-2}x^{n-r+2}$

Now, As given in the question,

$T_{r-1}:T_r:T_{r+1}=1:3:5$

$^nC_{r-2}:^nC_{r-1}:^nC_{r}=1:3:5$

$\frac{n!}{(r-2)!(n-r+2)!}:\frac{n!}{(r-1)!(n-r+1)!}:\frac{n!}{r!(n-r)!}=1:3:5$

From here, we get ,

$\frac{r-1}{n-r+2}=\frac{1}{3}\:\:and\:\:\frac{r}{n-r+1}=\frac{3}{5}$

Which can be written as

$n-4r+5=0\:\:and\:\:3n-8r+3=0$

From these equations we get,

$n=7\:\:and\:\:r=3$

As we know that the general  $(r+1)^{th}$ term  $T_{r+1}$ in the binomial expansion of  $(a+b)^n$  is given by

$T_{r+1}=^nC_ra^{n-r}b^r$

So, general  $(r+1)^{th}$ term  $T_{r+1}$ in the binomial expansion of  $(1+x)^{2n}$ is,

$T_{r+1}=^{2n}C_r1^{2n-r}x^r$

$x^n$ will come when $r=n$,

So, Coefficient of $x^n$ in the binomial expansion of  $(1+x)^{2n}$ is,

$K_{1x^n}=^{2n}C_n$

Now,

the general  $(r+1)^{th}$ term  $T_{r+1}$ in the binomial expansion of  $(1+x)^{2n-1}$ is,

$T_{r+1}=^{2n-1}C_r1^{2n-1-r}x^r$

Here also $x^n$ will come when $r=n$,

So, Coefficient of $x^n$ in the binomial expansion of  $(1+x)^{2n-1}$ is,

$K_{2x^n}=^{2n-1}C_n$

Now, As we can see

$^{2n-1}C_n=\frac{(2n-1)!}{n!(2n-1-n)!}=\frac{(2n-1)!}{n!(n-1)!}=\frac{(2n)!}{2n(n!)(n-1)!}=\frac{(2n)!}{2(n!)(n!)}$

$^{2n-1}C_n=\frac{1}{2}\times^{2n}C_n$

$2\times^{2n-1}C_n=^{2n}C_n$

$2\times K_{2x^n}=K_{1x^n}$

Hence, the coefficient of $x^n$ in the expansion of $(1+x)^{2n}$ is twice the coefficient of $x^n$ in the expansion of $(1+x)^{2n-1}$.

As we know that the general  $(r+1)^{th}$ term  $T_{r+1}$ in the binomial expansion of  $(a+b)^n$  is given by

$T_{r+1}=^nC_ra^{n-r}b^r$

So, the general  $(r+1)^{th}$ term  $T_{r+1}$ in the binomial expansion of  $(1 + x)^ m$  is

$T_{r+1}=^mC_r1^{m-r}x^r=^mC_rx^r$

$x^2$ will come when $r=2$. So,

The coeficient of $x^2$  in the binomial expansion of  $(1 + x)^ m$  = 6

$\Rightarrow ^mC_2=6$

$\Rightarrow \frac{m!}{2!(m-2)!}=6$

$\Rightarrow \frac{m(m-1)}{2}=6$

$\Rightarrow m(m-1)=12$

$\Rightarrow m^2-m-12=0$

$\Rightarrow (m+3)(m-4)=0$

$\Rightarrow m=4\:or\:-3$

Hence the positive value of m for which the coefficient of $x^2$ in the expansion $(1 + x)^ m$ is 6, is 4.

CBSE NCERT solutions for class 11 maths chapter 8 binomial theorem-Miscellaneous Exercise

As we know the Binomial expansion of $(a + b)^n$ is given by

$(a+b)^n=^nC_0a^n+^nC_1a^{n-1}b+^nC_2a^{n-2}b^2+......^nC_nb^n$

Given in the question,

$^nC_0a^n=729.......(1)$

$^nC_1a^{n-1}b=7290.......(2)$

$^nC_2a^{n-2}b^2=30375.......(3)$

Now, dividing (1) by (2) we get,

$\Rightarrow \frac{^nC_0a^n}{^nC_1a^{n-1}b}=\frac{729}{7290}$

$\Rightarrow \frac{\frac{n!}{n!0!}}{\frac{n!}{1!(n-1)!}}\times\frac{a}{b}=\frac{729}{7290}$

$\Rightarrow\frac{(n-1)!}{n!}\times\frac{a}{b}=\frac{1}{10}$

$\Rightarrow\frac{1}{n}\times\frac{a}{b}=\frac{1}{10}$

$10a=nb......(4)$

Now, Dividing (2) by (3) we get,

$\Rightarrow \frac{^nC_1a^{n-1}b}{^nC_2a^{n-2}b^2}=\frac{7290}{30375}$

$\Rightarrow \frac{\frac{n!}{1!(n-1)!}}{\frac{n!}{2!(n-2)!}}\times\frac{a}{b}=\frac{7290}{30375}$

$\Rightarrow \frac{2(n-2)!}{(n-1)!}\times\frac{a}{b}=\frac{7290}{30375}$

$\Rightarrow \frac{2}{(n-1)}\times\frac{a}{b}=\frac{7290}{30375}$

$\Rightarrow 2\times30375\times a=7290\times b\times(n-1)$

$\Rightarrow 60750a=7290b(n-1).......(5)$

Now, From (4) and (5), we get,

$n=6,a=3\:and\:b=5$

As we know that the general  $(r+1)^{th}$ term  $T_{r+1}$ in the binomial expansion of  $(a+b)^n$  is given by

$T_{r+1}=^nC_ra^{n-r}b^r$

So, the general  $(r+1)^{th}$ term  $T_{r+1}$ in the binomial expansion of  $(3 + ax)^9$  is

$T_{r+1}=^nC_r3^{n-r}(ax)^r=^nC_r3^{n-r}a^rx^r$

Now, $x^2$ will come when $r=2$ and  $x^3$ will come when $r=3$

So, the coefficient of $x^2$ is

$K_{x^2}=^nC_23^{9-2}a^2=^nC_23^7a^2$

And the coefficient of $x^3$ is

$K_{x^3}=^9C_33^{9-3}a^2=^9C_33^6a^3$

Now, Given in the question,

$K_{x^2}=K_{x^3}$

$^9C_23^7a^2=^9C_33^6a^3$

$\frac{9!}{2!7!}\times3=\frac{9!}{3!6!}\times a$

$a=\frac{18}{14}=\frac{9}{7}$

Hence the value of a is 9/7.

First, lets expand both expressions individually,

So,

$(1+2x)^6=^6C_0+^6C_1(2x)+^6C_2(2x)^2+^6C_3(2x)^3+^6C_4(2x)^4+^6C_5(2x)^5+$$^6C_6(2x)^6$

$(1+2x)^6=^6C_0+2\times^6C_1x+4\times^6C_2x^2+8\times^6C_3x^3+16\times^6C_4x^4+32\times^6C_5x^5+$$64\times^6C_6x^6$

$(1+2x)^6=1+12x+60x^2+160x^3+240x^4+192x^5+64x^6$

And

$(1-x)^7=^7C_0-^7C_1x+^7C_2x^2-^7C_3x^3+^7C_4x^4-^7C_5x^5+^7C_6x^6-^7C_7x^7$

$(1-x)^7=1-7x+21x^2-35x^3+35x^4-21x^5+7x^6-x^7$

Now,

$(1 + 2x)^6 (1 - x)^7=(1+12x+60x^2+160x^3+240x^4+192x^5+64x^6)$$(1-7x+21x^2-35x^3+35x^4-21x^5+7x^6-x^7)$

Now, for the coefficient of $x^5$, we multiply and add those terms whose product gives $x^5$.So,

The term which contain $x^5$are,

$\Rightarrow (1)(-21x^5)+(12x)(35x^4)+(60x^2)(-35x^3)+(160x^3)(21x^2)+(240x^4)(-7x)$$+(192x^5)(1)$

$\Rightarrow 171x^5$

Hence the coefficient of $x^5$ is 171.

we need to prove,

$a^n-b^n=k(a-b)$  where k is some natural number.

Now let's add and subtract b from a so that we can prove the above result,

$a=a-b+b$

$a^n=(a-b+b)^n=[(a-b)+b]^n$

$=^nC_0(a-b)^n+^nC_1(a-b)^{n-1}b+........^nC_nb^n$

$=(a-b)^n+^nC_1(a-b)^{n-1}b+........^nC_{n-1}(a-b)b^{n-1}+b^n$$\Rightarrow a^n-b^n=(a-b)[(a-b)^{n-1}+^nC_2(a-b)^{n-2}+........+^nC_{n-1}b^{n-1}]$

$\Rightarrow a^n-b^n=k(a-b)$

Hence,$a - b$ is a factor of $a^n - b^n$.

First let's simplify the expression $(a+b)^6-(a-b)^6$ using binomial theorem,

So,

$(a+b)^6=^6C_0a^6+^6C_1a^5b+^6C_2a^4b^2+^6C_3a^3b^3+^6C_4a^2b^4+^6C_5ab^5+^6C_6b^6$$(a+b)^6=a^6+6a^5b+15a^4b^2+20a^3b^3+15a^2b^4+6ab^5+b^6$

And

$(a-b)^6=^6C_0a^6-^6C_1a^5b+^6C_2a^4b^2-^6C_3a^3b^3+^6C_4a^2b^4-^6C_5ab^5+^6C_6b^6$

$(a+b)^6=a^6-6a^5b+15a^4b^2-20a^3b^3+15a^2b^4-6ab^5+b^6$

Now,

$(a+b)^6-(a-b)^6=a^6+6a^5b+15a^4b^2+20a^3b^3+15a^2b^4+6ab^5+b^6$$-a^6+6a^5b-15a^4b^2+20a^3b^3-15a^2b^4+6ab^5-b^6$

$(a+b)^6-(a-b)^6=2[6a^5b+20a^3b^3+6ab^5]$

Now, Putting $a=\sqrt{3}\:and\:b=\sqrt{2},$ we get

$\left(\sqrt3 + \sqrt2 \right )^6 - \left(\sqrt{3} - \sqrt2 \right )^6=2[6(\sqrt{3})^5(\sqrt{2})+20(\sqrt{3})^3(\sqrt{2})^3+6(\sqrt{3})(\sqrt{2})^5]$

$\left(\sqrt3 + \sqrt2 \right )^6 - \left(\sqrt{3} - \sqrt2 \right )^6=2[54\sqrt{6}+120\sqrt{6}+24\sqrt{6}]$

$\left(\sqrt3 + \sqrt2 \right )^6 - \left(\sqrt{3} - \sqrt2 \right )^6=2\times198\sqrt{6}$

$\left(\sqrt3 + \sqrt2 \right )^6 - \left(\sqrt{3} - \sqrt2 \right )^6=396\sqrt{6}$

First, lets simplify the expression $(x+y)^4-(x-y)^4$ using binomial expansion,

$(x+y)^4=^4C_0x^4+^4C_1x^3y+^4C_2x^2y^2+^4C_3xy^3+^4C_4y^4$

$(x+y)^4=x^4+4x^3y+6x^2y^2+4xy^3+y^4$

And

$(x-y)^4=^4C_0x^4-^4C_1x^3y+^4C_2x^2y^2-^4C_3xy^3+^4C_4y^4$

$(x-y)^4=x^4-4x^3y+6x^2y^2-4xy^3+y^4$

Now,

$(x+y)^4-(x-y)^4=x^4+4x^3y+6x^2y^2+4xy^3+y^4-$$x^4+4x^3y-6x^2y^2+4xy^3-y^4$

$(x+y)^4-(x-y)^4=2(x^4+6x^2y^2+y^4)$

Now, Putting $x=a^2\and\:y=\sqrt{a^2-1}$ we get,

$\left(a^2 + \sqrt{a^2 -1} \right )^4 + \left(a^2 - \sqrt{a^2 -1} \right )^4=2[(a^2)^4+6(a^2)^2(\sqrt{a^2-1})^2+(\sqrt{a^2-1})^4]$

$\left(a^2 + \sqrt{a^2 -1} \right )^4 + \left(a^2 - \sqrt{a^2 -1} \right )^4=2[a^8+6a^4(a^2-1)+(a^2-1)^2]$

$\left(a^2 + \sqrt{a^2 -1} \right )^4 + \left(a^2 - \sqrt{a^2 -1} \right )^4=2a^8+12a^6-12a^4+2a^4-4a^2+2$

$\left(a^2 + \sqrt{a^2 -1} \right )^4 + \left(a^2 - \sqrt{a^2 -1} \right )^4=2a^8+12a^6-10a^4-4a^2+2$

As we can write 0.99 as 1-0.01,

$(0.99)^5=(1-0.001)^5=^5C_0(1)^5-^5C_1(1)^4(0.01)+^5C_2(1)^3(0.01)^2$$+\:other \:negligible \:terms$

$\Rightarrow (0.99)^5=1-5(0.01)+10(0.01)^2$

$\Rightarrow (0.99)^5=1-0.05+0.001$

$\Rightarrow (0.99)^5=0.951$

Hence the value of   $(0.99)^5$  is 0.951 approximately.

Given, the expression

$\left(\sqrt[4]{2} + \frac{1}{\sqrt[4]{3}} \right )^n$

Fifth term from the beginning  is

$T_5=^nC_4(\sqrt[4]{2})^{n-4}\left(\frac{1}{\sqrt[4]{3}}\right)^4$

$T_5=^nC_4\frac{(\sqrt[4]{2})^n}{(\sqrt[4]{2})^4}\times\frac{1}{3}$

$T_5=\frac{n!}{4!(n-4)!}\times\frac{(\sqrt[4]{2})^n}{2}\times\frac{1}{3}$

And Fifth term from the end is,

$T_{n-5}=^nC_{n-4}(\sqrt[4]{2})^4\left ( \frac{1}{\sqrt[4]{3}} \right )^{n-4}$

$T_{n-5}=^nC_{n-4}(\sqrt[4]{2})^4\left ( \frac{(\sqrt[4]{3})^4}{(\sqrt[4]{3})^n} \right )$

$T_{n-5}=\frac{n!}{4!(n-4)!}\times2\times\left ( \frac{3}{(\sqrt[4]{3})^n} \right )$

Now, As given in the question,

$T_5:T_{n-5}=\sqrt{6}:1$

So,

$\left(\frac{n!}{4!(n-4)!}\times\frac{(\sqrt[4]{2})^n}{2}\times\frac{1}{3}\right):\left( \frac{n!}{4!(n-4)!}\times2\times\left ( \frac{3}{(\sqrt[4]{3})^n} \right )\right)=\sqrt{6}:1$

From Here ,

$\frac{(\sqrt[4]{2})^n}{6}:\frac{6}{(\sqrt[4]{3})^n}=\sqrt{6}:1$

$\frac{(\sqrt[4]{2})^n(\sqrt[4]{3})^n}{6\times6}=\sqrt{6}$

$(\sqrt[4]{6})^n=36\sqrt{6}$

$6^{\frac{n}{4}}=6^{\frac{5}{2}}$

From here,

$\frac{n}{4}=\frac{5}{2}$

$n=10$

Hence the value of n is 10.

Given the expression,

$\left(1 + \frac{x}{2} - \frac{2}{x} \right )^4, \ x\neq 0$

Binomial expansion of this expression is

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

$\Rightarrow \left ( 1+\frac{x}{2} \right )^4-\frac{8}{x}\left ( 1+\frac{x}{2} \right )^3+$$\frac{24}{x^2}+\frac{24}{x}+6$$-\frac{32}{x^3}+\frac{16}{x^4}..........(1)$

Now Applying Binomial Theorem again,

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

$= 1+ 4\left ( \frac{x}{2} \right )+6\left ( \frac{x^2}{4} \right )+4\left ( \frac{x^3}{8} \right )+\frac{x^4}{16}$

$=1+2x+\frac{3x^2}{2}+\frac{x^3}{3}+\frac{x^4}{16}..............(2)$

And

$\left ( 1+\frac{x}{2} \right )^3=^3C_0(1)^3+^3C_1(1)^2\left ( \frac{x}{2} \right )+^3C_2(1)\left ( \frac{x}{2} \right )^2+^3C_3\left ( \frac{x}{2} \right )^3$

$\left ( 1+\frac{x}{2} \right )^3= 1+\frac{3x}{2}+\frac{3x^2}{4}+\frac{x^3}{8}..........(3)$

Now, From (1), (2) and (3) we get,

$\\\left(1 + \frac{x}{2} - \frac{2}{x} \right )^4=1+2x+\frac{3x^2}{2}+\frac{x^3}{8}+\frac{x^4}{16}-\frac{8}{x}\left ( 1+\frac{3x}{2}+\frac{3x^2}{4}+\frac{x^2}{8} \right )$$+\frac{8}{x^2}+\frac{24}{x}+6-\frac{32}{x^3}+\frac{16}{x^4}$

$\\\left(1 + \frac{x}{2} - \frac{2}{x} \right )^4=1+2x+\frac{3x^2}{2}+\frac{x^3}{8}+\frac{x^4}{16}-\frac{8}{x}-12-6x-x^2$$+\frac{8}{x^2}+\frac{24}{x}+6-\frac{32}{x^3}+\frac{16}{x^4}$

$\\\left(1 + \frac{x}{2} - \frac{2}{x} \right )^4=1+2x+\frac{3x^2}{2}+\frac{x^3}{8}+\frac{x^4}{16}-\frac{8}{x}\left ( 1+\frac{3x}{2}+\frac{3x^2}{4}+\frac{x^2}{8} \right )$$=\frac{16}{x}+\frac{8}{x^2}-\frac{32}{x^3}+\frac{16}{x^4}-4x+\frac{x^2}{2}+\frac{x^3}{2}+\frac{x^4}{16}-5$

Given $(3x^2 - 2ax +3a^2)^3$

By Binomial Theorem It can also be written as

$(3x^2 - 2ax +3a^2)^3=((3x^2 - 2ax) +3a^2)^3$

$= ^3C_0(3x^2-2ax)^3+^3C_1(3x^2-2ax)^2(3a^2)+^3C_2(3x^2-2ax)(3a^2)^2+^3C_3(3a^2)^3$

$= (3x^2-2ax)^3+3(3x^2-2ax)^2(3a^2)+3(3x^2-2ax)(3a^2)^2+(3a^2)^3$

$= (3x^2-2ax)^3+81a^2x^4-108a^3x^3+36a^4x^2+81a^4x^2-54a^5x+27a^6$

$= (3x^2-2ax)^3+81a^2x^4-108a^3x^3+117a^4x^2-54a^5x+27a^6...........(1)$

Now, Again By Binomial Theorem,

$(3x^2-2ax)^3=^3C_0(3x^2)^3-^3C_1(3x^2)^2(2ax)+^3C_2(3x^2)(2ax)^2-^3C_3(2ax)^3$

$(3x^2-2ax)^3= 27x^6-3(9x^4)(2ax)+3(3x^2)(4a^2x^2)-8a^2x^3$

$(3x^2-2ax)^3= 27x^6-54x^5+36a^2x^4-8a^3x^3............(2)$

From (1) and (2) we get,

$(3x^2-2ax+3a^2)^3=27x^6-54x^5+36a^2x^3+81a^2x^4-108a^3x^3+117a^4x^2$$-54a^5x+27a^6$

$(3x^2-2ax+3a^2)^3=27x^6-54x^5+117a^2x^3-116a^3x^3+117a^4x^2$$-54a^5x+27a^6$

## NCERT solutions for class 11 mathematics

 chapter-1 NCERT solutions for class 11 maths chapter 1 Sets chapter-2 Solutions of NCERT for class 11 chapter 2 Relations and Functions chapter-3 CBSE NCERT solutions for class 11 chapter 3 Trigonometric Functions chapter-4 NCERT solutions for class 11 chapter 4 Principle of Mathematical Induction chapter-5 Solutions of NCERT for class 11 chapter 5 Complex Numbers and Quadratic equations chapter-6 CBSE NCERT solutions for class 11 maths chapter 6 Linear Inequalities chapter-7 NCERT solutions for class 11 maths chapter 7 Permutation and Combinations chapter-8 NCERT solutions for class 11 maths chapter 8 Binomial Theorem chapter-9 CBSE NCERT solutions for class 11 maths chapter 9 Sequences and Series chapter-10 NCERT solutions for class 11 maths chapter 10 Straight Lines chapter-11 Solutions of NCERT for class 11 maths chapter 11 Conic Section chapter-12 CBSE NCERT solutions for class 11 maths chapter 12 Introduction to Three Dimensional Geometry chapter-13 NCERT solutions for class 11 maths chapter 13 Limits and Derivatives chapter-14 Solutions of NCERT for class 11 maths chapter 14 Mathematical Reasoning chapter-15 CBSE NCERT solutions for class 11 maths chapter 15 Statistics chapter-16 NCERT solutions for class 11 maths chapter 16 Probability

## NCERT solutions for class 11- Subject wise

 Solutions of NCERT for class 11 biology CBSE NCERT solutions for class 11 maths NCERT solutions for class 11 chemistry Solutions of NCERT for Class 11 physics

In NCERT solutions for class 11 maths chapter 8 binomial theorem, there are some important formulas to be remembered which are mentioned below.

The binomial theorem for a positive integer n

$(a+b)^{n}=^{n} \mathrm{C}_{0} a^{n}+^{n} \mathrm{C}_{1} a^{n-1} b+^{n} \mathrm{C}_{2} a^{n-2} b^{2}+\ldots+^{n} \mathrm{C}_{n-1} a \cdot b^{n-1}+^{n} \mathrm{C}_{n} b^{n}$

$(a+b)^{n}=\sum_{k=0}^{n}^{n} \mathrm{C}_{k} a^{n-k} b^{k}$

$^{n} \mathrm{C}_{r}$ -> binomial coefficients.

Some special cases

Put a=1, b=x     $(1+x)^{n}=^{n} C_{0}+^{n} C_{1} x+^{n} C_{z} x^{2}+^{n} C_{3} x^{3}+\ldots+^{n} C_{n} x^{n}$

Put x=1             $2^{n}=^{n} \mathrm{C}_{0}+^{n} \mathrm{C}_{1}+^{n} \mathrm{C}_{2}+\ldots+^{n} \mathrm{C}_{n}$

Put a=1,b=-x     $(1-x)^{n}=^{n} \mathrm{C}_{0}-^{n} \mathrm{C}_{1} x+^{n} \mathrm{C}_{2} x^{2}-\ldots+(-1)^{n} \mathrm{C}_{n} x^{n}$

Put x=1             $0=^{n} \mathrm{C}_{0}-^{n} \mathrm{C}_{1}+^{n} \mathrm{C}_{2}-\ldots+(-1)^{n} \mathrm{C}_{n}$

There are 10 problems in miscellaneous exercise. To get command on this chapter, you need to solve miscellaneous exercise too. In NCERT solutions for class 11 maths chapter 8 binomial theorem, you will get solutions to miscellaneous exercise too.