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.

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.

 

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

\Rightarrow64x^6-6(32x^5)(3)+15(16x^4)(9)-20(8x^3)(27)+15(4x^2)(81)-6(2x)(243)+729

\Rightarrow64x^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 - 9is divisible by 64, whenever n is a positive integer.

Answer:

If we want to prove that  9^{n+1} - 8n - 9is 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 4th 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 13th 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 , rth 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.

CBSE 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^2and 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^5are,

\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 thata - 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 )^nis \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

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. 

Happy Reading !!! 

 

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