NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem

Q: What are important topics of the chapter Chapter 8 Binomial Theorem ?

Basic concept of binomial theorem, b inomial theorem for positive integral indices, g eneral and middle terms are covered in this chapter.

Updated on Aug 4, 2020 - 10:07 p.m. IST by Ravindra Pindel

NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem: You have studied the expansion of expressions like (a-b) ² and (a-b) ³ 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 NCERT solutions 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. Here you will get NCERT solutions for class 11 also.

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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 0.99 ⁶ using the first three terms of expansion? This can be solved by rewriting 0.99 ⁶ as (1-0.01) ⁶ and expanding using the Binomial Theorem.

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The NCERT Solutions of this chapter are given below:

NCERT solutions for class 11 maths chapter 8 binomial theorem-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 for class 11 maths chapter 8 binomial theorem-Exercise: 8.2

Question:1 Find the coefficient of

$x^5$ in $(x + 3)^8$

Answer:

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

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

Question:2 Find the coefficient of $a^5b^7$ in $(a- 2b)^{12}$

Answer:

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

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

Question:3 Write the general term in the expansion of

$(x^2 - y)^6$

Answer:

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 $(x^2 - y)^6$ :

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

Question:4 Write the general term in the expansion of

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

Answer:

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

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

Question:5 Find the 4 ^th term in the expansion of $(x-2y)^{12}$ .

Answer:

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

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

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

Question:6 Find the 13 ^th term in the expansion of $\left(9x - \frac{1}{3\sqrt x} \right )^{18},\ x\neq 0$

Answer:

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

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

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

$=18564$

Question:7 Find the middle terms in the expansion of $\left(3 - \frac{x^3}{6} \right )^7$

Answer:

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

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

Question:8 Find the middle terms in the expansion of $\left(\frac{x}{3} + 9y \right )^{10}$

Answer:

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; $61236x^5y^5$ .

Question:9 In the expansion of $(1 + a)^{m+n}$ , prove that coefficients of $a^m$ and $a^n$ are equal

Answer:

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.

Question:10 The coefficients of the $(r-1)$ ^th , r ^th and $(r + 1)$ ^th terms in the expansion of $(x+1)^{n}$ are in the ratio 1 : 3 : 5. Find n and r .

Answer:

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,

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

Question:11 Prove that the coefficient of $x^n$ in the expansion of $(1+x)^{2n}$ is twice the coefficient of $x^n$ in the expansion of $(1+x)^{2n-1}$ .

Answer:

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 $(1+x)^{2n-1}$ .

Question:12 Find a positive value of m for which the coefficient of $x^2$ in the expansion $(1 + x)^ m$ is 6.

Answer:

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.

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

Question:1 Find a , b and n in the expansion of $(a + b)^n$ if the first three terms of the expansion are 729, 7290 and 30375, respectively.

Answer:

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,

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

Question:2 Find a if the coefficients of $x^2$ and $x^3$ in the expansion of $(3 + ax)^9$ are equal.

Answer:

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.

Question:3 Find the coefficient of $x^5$ in the product $(1 + 2x)^6 (1 - x)^7$ using binomial theorem.

Answer:

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.

Question:4 If a and b are distinct integers, prove that $a - b$ is a factor of $a^n - b^n$ , whenever n is a positive integer.
[ Hint: write $a^n = (a - b + b)^n$ and expand]

Answer:

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 $a^n - b^n$ .

Question:5 Evaluate $\left(\sqrt3 + \sqrt2 \right )^6 - \left(\sqrt{3} - \sqrt2 \right )^6$ .

Answer:

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

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

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

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

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

Question:6 Find the value of $\left(a^2 + \sqrt{a^2 -1} \right )^4 + \left(a^2 - \sqrt{a^2 -1} \right )^4$

Answer:

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,

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

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

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

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

Question:7 Find an approximation of (0.99) ⁵ using the first three terms of its expansion.

Answer:

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.

Question:8 Find n, if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of $\left(\sqrt[4]{2} + \frac{1}{\sqrt[4]{3}} \right )^n$ is $\sqrt6 :1$

Answer:

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.

Question:9 Expand using Binomial Theorem $\left(1 + \frac{x}{2} - \frac{2}{x} \right )^4, \ x\neq 0$

Answer:

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,

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

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

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

Question:10 Find the expansion of $(3x^2 - 2ax +3a^2)^3$ using binomial theorem .

Answer:

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	NCERT solutions for class 11 chapter 2 Relations and Functions
chapter-3	NCERT solutions for class 11 chapter 3 Trigonometric Functions
chapter-4	NCERT solutions for class 11 chapter 4 Principle of Mathematical Induction
chapter-5	NCERT solutions for class 11 chapter 5 Complex Numbers and Quadratic equations
chapter-6	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	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	NCERT solutions for class 11 maths chapter 11 Conic Section
chapter-12	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	NCERT solutions for class 11 maths chapter 14 Mathematical Reasoning
chapter-15	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

NCERT solutions for class 11 biology

NCERT solutions for class 11 maths

NCERT solutions for class 11 chemistry

NCERT solutions 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.

Happy Reading !!!

Solutions

Frequently Asked Question (FAQs) - NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem

Question: What are important topics of the chapter Chapter 8 Binomial Theorem ?

Answer:

Basic concept of binomial theorem, binomial theorem for positive integral indices, general and middle terms are covered in this chapter.

Question: How does the NCERT solutions are helpful ?

Answer:

NCERT solutions are helpful to the students while solving the NCERT problems. If they are stuck while solving, they can take help these solutions provided provided in a detailed manner.

Question: What is the weightage of chapter Binomial Theorem in Jee main ?

Answer:

Binomial Theorem has 4 % weightage in the Jee main exam.

Question: Where can I find the complete solutions of NCERT for class 11 maths ?

Answer:

Here you will get the detailed NCERT solutions for class 11 maths by clicking on the link.

Question: What is the weightage of chapter Binomial Theorem in Jee advanced ?

Answer:

Binomial Theorem has about 3 % weightage in the Jee main exam.

Question: Which is the official website of NCERT ?

Answer:

NCERT official is the official website of the NCERT where you can get NCERT textbooks and syllabus from class 1 to 12.

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NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem

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

NCERT solutions for class 11 maths chapter 8 binomial theorem-Exercise: 8.2

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

Frequently Asked Question (FAQs) - NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem

Question: What are important topics of the chapter Chapter 8 Binomial Theorem ?

Question: How does the NCERT solutions are helpful ?

Question: What is the weightage of chapter Binomial Theorem in Jee main ?

Question: Where can I find the complete solutions of NCERT for class 11 maths ?

Question: What is the weightage of chapter Binomial Theorem in Jee advanced ?

Question: Which is the official website of NCERT ?

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