NCERT Solutions for Class 11 Maths Chapter 7 Exercise 7.1 - Binomial Theorem

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

Answer:

Given,

The Expression:

$(1-2x{)}^5$

the expansion of this Expression is,

$(1-2x{)}^5=$

$$\begin{gathered} { \\ }^5{C}_0(1{)}^5{-}^5{C}_1(1{)}^4(2x){+}^5{C}_2(1{)}^3(2x{)}^2{-}^5{C}_3(1{)}^2(2x{)}^3{+}^5{C}_4(1{)}^1(2x{)}^4{-}^5{C}_5(2x{)}^5 \end{gathered}$$

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

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

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

Answer:

Given,

The Expression:

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

the expansion of this Expression is,

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

$$\begin{gathered} { \\ }^5{C}_0{\left(\frac{2}{x}\right)}^5{-}^5{C}_1{\left(\frac{2}{x}\right)}^4\left(\frac{x}{2}\right){+}^5{C}_2{\left(\frac{2}{x}\right)}^3{\left(\frac{x}{2}\right)}^2 \end{gathered}$$${-}^5{C}_3{\left(\frac{2}{x}\right)}^2{\left(\frac{x}{2}\right)}^3{+}^5{C}_4{\left(\frac{2}{x}\right)}^1{\left(\frac{x}{2}\right)}^4{-}^5{C}_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$$$\begin{gathered} { \\ }^6{C}_0(2x{)}^6{-}^6{C}_1(2x{)}^5(3){+}^6{C}_2(2x{)}^4(3{)}^2{-}^6{C}_3(2x{)}^3(3{)}^3+ \end{gathered}$$${}^6{C}_4(2x{)}^2(3{)}^4{-}^6{C}_5(2x)(3{)}^5{+}^6{C}_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. ${(\frac{x}{3}+\frac{1}{x})}^5$

Answer:

Given,

The Expression:

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

the expansion of this Expression is,

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

${\Rightarrow }^5{C}_0{\left(\frac{x}{3}\right)}^5{+}^5{C}_1{\left(\frac{x}{3}\right)}^4\left(\frac{1}{x}\right){+}^5{C}_2{\left(\frac{x}{3}\right)}^3{\left(\frac{1}{x}\right)}^2$${+}^5{C}_3{\left(\frac{x}{3}\right)}^2{\left(\frac{1}{x}\right)}^3{+}^5{C}_4{\left(\frac{x}{3}\right)}^1{\left(\frac{1}{x}\right)}^4{+}^5{C}_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. ${(x+\frac{1}{x})}^6$

Answer:

Given,

The Expression:

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

the expansion of this Expression is,

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

${\Rightarrow }^6{C}_0(x{)}^6{+}^6{C}_1(x{)}^5\left(\frac{1}{x}\right){+}^6{C}_2(x{)}^4{\left(\frac{1}{x}\right)}^2{+}^6{C}_3(x{)}^3{\left(\frac{1}{x}\right)}^3+$${}^6{C}_4(x{)}^2{\left(\frac{1}{x}\right)}^4{+}^6{C}_5(x){\left(\frac{1}{x}\right)}^5{+}^6{C}_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);

$$\begin{gathered} \Rightarrow (96{)}^3 \\ =(100-4{)}^3 \\ {=}^3{C}_0(100{)}^3{-}^3{C}_1(100{)}^2(4){+}^3{C}_2(100)(4{)}^2{-}^3{C}_3(4{)}^3 \end{gathered}$$

$=(100{)}^3-3(100{)}^2(4)+3(100)(4{)}^2-(4{)}^3$

$=1000000-120000+4800-64$

$=884736$

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

Answer:

As we can write 102 in the form 100+2

$\Rightarrow (102{)}^5$

$=(100+2{)}^5$

${=}^5{C}_0(100{)}^5{+}^5{C}_1(100{)}^4(2){+}^5{C}_2(100{)}^3(2{)}^2$${+}^5{C}_3(100{)}^2(2{)}^3{+}^5{C}_4(100{)}^1(2{)}^4{+}^5{C}_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

$\Rightarrow (101{)}^4$

$=(100+1{)}^4$

${=}^4{C}_0(100{)}^4{+}^4{C}_1(100{)}^3(1){+}^4{C}_2(100{)}^2(1{)}^2$${+}^4{C}_3(100{)}^1(1{)}^3{+}^4{C}_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

$\Rightarrow (99{)}^5$

$=(100-1{)}^5$

${=}^5{C}_0(100{)}^5{-}^5{C}_1(100{)}^4(1){+}^5{C}_2(100{)}^3(1{)}^2$${-}^5{C}_3(100{)}^2(1{)}^3{+}^5{C}_4(100{)}^1(1{)}^4{-}^5{C}_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,

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

${=}^{10000}{C}_0{+}^{10000}{C}_1(1.1)+Otherpositiveterms$

$=1+10000\times 1.1+Otherpositiveterm$

$>1000$

Hence,

$(1.1{)}^{10000}>1000$

Question 11 :Find $(a+b{)}^4-(a-b{)}^4$ . Hence, evaluate$(\sqrt{3}+\sqrt{2}{)}^4-(\sqrt{3}-\sqrt{2}{)}^4$ .

Answer:

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

$(a+b{)}^4{=}^4{C}_0{a}^4{+}^4{C}_1{a}^{3}b{+}^4{C}_2{a}^2{b}^2{+}^4{C}_{3}a{b}^3{+}^4{C}_4{b}^4$

$(a-b{)}^4{=}^4{C}_0{a}^4{-}^4{C}_1{a}^{3}b{+}^4{C}_2{a}^2{b}^2{-}^4{C}_{3}a{b}^3{+}^4{C}_4{b}^4$

From Here,

$(a+b{)}^4-(a-b{)}^4{=}^4{C}_0{a}^4{+}^4{C}_1{a}^{3}b{+}^4{C}_2{a}^2{b}^2{+}^4{C}_{3}a{b}^3{+}^4{C}_4{b}^4$${-}^4{C}_0{a}^4{+}^4{C}_1{a}^{3}b{-}^4{C}_2{a}^2{b}^2{+}^4{C}_{3}a{b}^3{-}^4{C}_4{b}^4$

$(a+b{)}^4-(a-b{)}^4=2\times {(}^4{C}_1{a}^{3}b{+}^4{C}_{3}a{b}^3)$

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

Now, Using this, we get

$(\sqrt{3}+\sqrt{2}{)}^4-(\sqrt{3}-\sqrt{2}{)}^4=8(\sqrt{3})(\sqrt{2})(3+2)=8\times \sqrt{6}\times 5=40\sqrt{6}$

Question 12 :Find $(x+1{)}^6+(x-1{)}^6$ . Hence or otherwise evaluate $(\sqrt{2}+1{)}^6+(\sqrt{2}-1{)}^6$.

Answer:

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

$(x+1{)}^6{=}^6{C}_0{x}^6{+}^6{C}_1{x}^{5}1{+}^6{C}_2{x}^4{1}^2{+}^4{C}_3{x}^3{1}^3{+}^6{C}_4{x}^2{1}^4{+}^6{C}_{5}x{1}^5{+}^6{C}_6{1}^6$

$(x-1{)}^6{=}^6{C}_0{x}^6{-}^6{C}_1{x}^{5}1{+}^6{C}_2{x}^4{1}^2{-}^4{C}_3{x}^3{1}^3{+}^6{C}_4{x}^2{1}^4{-}^6{C}_{5}x{1}^5{+}^6{C}_6{1}^6$

From Here,

$(x+1{)}^6-(x-1{)}^6{=}^6{C}_0{x}^6{+}^6{C}_1{x}^{5}1{+}^6{C}_2{x}^4{1}^2{+}^4{C}_3{x}^3{1}^3+$${}^6{C}_4{x}^2{1}^4{+}^6{C}_{5}x{1}^5{+}^6{C}_6{1}^6$${+}^6{C}_0{x}^6{-}^6{C}_1{x}^{5}1{+}^6{C}_2{x}^4{1}^2{-}^4{C}_3{x}^3{1}^3{+}^6{C}_4{x}^2{1}^4{-}^6{C}_{5}x{1}^5{+}^6{C}_6{1}^6$

$(x+1{)}^6+(x-1{)}^6=2{(}^6{C}_0{x}^6{+}^6{C}_2{x}^4{1}^2{+}^6{C}_4{x}^2{1}^4{+}^6{C}_6{1}^6)$

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

Now, Using this, we get

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

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

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,

$(1+x{)}^m{=}^m{C}_0{+}^m{C}_{1}x{+}^m{C}_2{x}^2{+}^m{C}_3{x}^3+...{.}^m{C}_m{x}^m$

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

$9^{n+1}{=}^{n+1}{C}_0+{}^{n+1}{C}_{1}8{+}^{n+1}{C}_2{8}^2+.......{+}^{n+1}{C}_{n+1}{8}^{n+1}$

$9^{n+1}=1+8(n+1)+8^2{(}^{n+1}{C}_2+{}^{n+1}{C}_{3}8{+}^{n+1}{C}_4{8}^2+.......{+}^{n+1}{C}_{n+1}{8}^{n-1})$

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

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 $\sum _{r=0}^n{3}^r{}^n{C}_r=4^n$

Answer:

As we know from Binomial Theorem,

$\sum _{r=0}^n{a}^r{}^n{C}_r=(1+a{)}^n$

Here putting a = 3, we get,

$\sum _{r=0}^n{3}^r{}^n{C}_r=(1+3{)}^n$

$\sum _{r=0}^n{3}^r{}^n{C}_r=4^n$

Hence Proved.