NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem

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

The basic concept of the binomial theorem, b inomial theorem for positive integral indices, and g eneral and middle terms are covered in this chapter. students can go through the NCERT syllabus , all the important topics are mentioned there. practicing these topics covered in binomial theorem ncert solutions is crucial for commanding the concepts.

Edited By Ramraj Saini | Updated on Sep 23, 2023 06:17 PM IST

Binomial Theorem Class 11 Questions And Answers

NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem are provided here. These NCERT solutions are prepared by subjects matter experts considering the latest syllabus and pattern of CBSE 2023-24. You have studied the expansion of expressions like (a-b)²and (a-b)³in 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 class 11 maths chapter 8 NCERT solutions, you will get the answer to the above question. In this NCERT Book 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.

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Binomial Theorem Class 11 Questions And Answers
Binomial Theorem Class 11 Questions And Answers PDF Free Download
Binomial Theorem Class 11 Solutions - Important Formulae
Binomial Theorem Class 11 NCERT Solutions (Intext Questions and Exercise)
Highlights of NCERT Solutions for Class 11 Maths Chapter 8 – Binomial Theorem
Key Features of NCERT Solutions for Class 11 Maths Chapter 8 – Binomial Theorem

NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem

In NCERT solutions for class 11 chapter 8 binomial theorem, you will get questions related to these topics. NCERT Book 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 the NCERT syllabus of this chapter, there are 26 problems in 2 exercises. All these questions are prepared in binomial theorem class 11 NCERT solutions 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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Binomial Theorem Class 11 Questions And Answers PDF Free Download

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Binomial Theorem Class 11 Solutions - Important Formulae

Binomial Theorem:

The Binomial Theorem provides the expansion of a binomial (a + b) raised to any positive integer n.

The expansion of (a + b)ⁿ is given by: (a + b)ⁿ = nC0 * aⁿ + nC1 * a^(n-1) * b + nC2 * a^(n-2) * b² + … + nCn-1 * a * b^(n-1) + nCn * bⁿ.

Special Cases from the Binomial Theorem:

(x - y)ⁿ = nC0 * xⁿ - nC1 * x^(n-1) * y + nC2 * x^(n-2) * y² + … + (-1)ⁿ * nCn * x.

(1 - x)ⁿ = nC0 - nC1 * x + nC2 * x² - …. + (-1)ⁿ * nCn * xⁿ.

The coefficients nC0 and nCn are always equal to 1.

Pascal’s Triangle:

The coefficients of the expansions in the Binomial Theorem are arranged in an array called Pascal’s triangle.

General Term of Expansions:

For (a + b)ⁿ, the general term is T_{r+1} = nCr * a^(n-r) * b^r.

For (a - b)ⁿ, the general term is (-1)r * nCr * a^(n-r) * b^r.

For (1 + x)ⁿ, the general term is nCr * x^r.

For (1 - x)ⁿ, the general term is (-1)^r * nCn * xⁿ.

Middle Terms:

In the expansion (a + b)ⁿ, if n is even, then the middle term is the (n/2 + 1)-th term.

If n is odd, then the middle terms are the (n/2 + 1)-th and ((n+1)/2 + 1)-th terms.

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Binomial Theorem Class 11 NCERT Solutions (Intext Questions and Exercise)

Binomial theorem class 11 solutions - 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.

Binomial theorem class 11 solutions - 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.

Class 11 maths chapter 8 question answer - 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,

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

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

Highlights of NCERT Solutions for Class 11 Maths Chapter 8 – Binomial Theorem

8.1 Introduction

8.2 Binomial Theorem for Positive Integral Indices

Pascal’s Triangle

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8.2.1 Binomial theorem for any positive integer n

8.2.2 Some special cases

8.3 General and Middle Term

If you are interested in Binomial Theorem class 11 exercise solutions then these are listed below.

Binomial Theorem class 11 exercise 8.1

Binomial Theorem class 11 exercise 8.2

Binomial Theorem class 11 exercise miscellaneous exercise

NCERT solutions for class 11 mathematics

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

Key Features of NCERT Solutions for Class 11 Maths Chapter 8 – Binomial Theorem

Comprehensive explanations: The solutions of maths chapter 8 class 11 provide step-by-step explanations to all the questions in the chapter, helping students to understand the concepts thoroughly.

NCERT-based: The solutions of ch 8 maths class 11 are strictly based on the latest NCERT syllabus and follow the guidelines set by the board.

Easy language: The class 11 maths ch 8 question answer are written in simple and easy-to-understand language, making it easy for students to grasp the concepts.

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Important Formulas

The binomial theorem for a positive integer n

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

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

$^{n} \mathrm{C}_{r}$ -> 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 with miscellaneous exercise. To get command on this chapter, you need to solve miscellaneous exercises too. In NCERT solutions for class 11 maths chapter 8 binomial theorem, you will get solutions to miscellaneous exercises too.

Also Check NCERT Books and NCERT Syllabus here

Happy Reading !!!

Frequently Asked Questions (FAQs)

1. What are important topics of the chapter Chapter 8 Binomial Theorem ?

The basic concept of the binomial theorem, binomial theorem for positive integral indices, and general and middle terms are covered in this chapter. students can go through the NCERT syllabus, all the important topics are mentioned there. practicing these topics covered in binomial theorem ncert solutions is crucial for commanding the concepts.

2. How does the NCERT solutions are helpful ?

NCERT solutions are helpful to the students while solving the NCERT problems. After solving NCERT book problems students can acquire command in concepts that will help greatly during premier exams. If they are stuck while solving, they can take help with binomial theorem class 11 NCERT solutions provided in a detailed manner. for ease students can study class 11 maths chapter 8 solutions pdf both online and offline mode.

3. Explain the concept of the Binomial Theorem covered in NCERT solutions for class 11th maths chapter 8.

The Binomial Theorem refers to the method of expanding the power of binomials that involve the addition of two or more terms. The coefficients of the terms in the expansion are known as binomial coefficients. This chapter provides essential definitions that are relevant for examinations. With the NCERT Solutions available in PDF format, students can stay up-to-date with the latest CBSE Board syllabus.

4. Where can I find the complete solutions of NCERT for class 11 maths ?

Here you will get the detailed NCERT solutions for class 11 maths by clicking on the link. if anyone facing problems to find complete solutions of NCERT Book can web through official website of Careers360 or these are listed above in the article according to topic and subject wise.

5. In the Binomial Theorem, explain the properties of positive integers covered in the class 11 maths NCERT solutions chapter 8.

The NCERT Solutions for Class 11 Maths Chapter 8 provide over 10 properties related to positive integers that students can learn. These properties are crucial in comprehending the efficient solution of equations. During the annual examination, the question paper may focus on simple chapters that are challenging to solve. Hence, it is recommended that students go through these NCERT Solutions to secure good marks in the examination.

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem

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Key Features of NCERT Solutions for Class 11 Maths Chapter 8 – Binomial Theorem

NCERT solutions for class 11- Subject wise

Important Formulas

The binomial theorem for a positive integer n

Also Check NCERT Books and NCERT Syllabus here

Frequently Asked Questions (FAQs)

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