NCERT Solutions for Class 11 Maths Chapter 7 Binomial Theorem

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,

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

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

$=\frac{32}{x}-5\left(\frac{16}{x^4}\right)\left(\frac{x}{2}\right)+10\left(\frac{8}{x^3}\right)\left(\frac{x^2}{4}\right)-10\left(\frac{4}{x^2}\right)\left(\frac{x^2}{8}\right)$ $+5\left(\frac{2}{x}\right)\left(\frac{x^4}{16}\right)-\frac{x^5}{32}$

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

Question:3 Expand the expression. $(2x-3{)}^6$

Answer:

Given,

The Expression:

$(2x-3{)}^6$

the expansion of this Expression is,

$(2x-3{)}^6$

${=}^6{C}_0(2x{)}^6{-}^6{C}_1(2x{)}^5(3){+}^6{C}_2(2x{)}^4(3{)}^2{-}^6{C}_3(2x{)}^3(3{)}^3+$ ${}^6{C}_4(2x{)}^2(3{)}^4{-}^6{C}_5(2x)(3{)}^5{+}^6{C}_6(3{)}^6$

$=64x^6-6(32x^5)(3)+15(16x^4)(9)-20(8x^3)(27)+15(4x^2)(81)-6(2x)(243)$ $+729$

$=64x^6-576x^5+2160x^4-4320x^3+4860x^2-2916x+729$

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,

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

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

$=\frac{x^5}{243}+5\left(\frac{x^4}{81}\right)\left(\frac{1}{x}\right)+10\left(\frac{x^3}{27}\right)\left(\frac{1}{x^2}\right)+10\left(\frac{x^2}{9}\right)\left(\frac{1}{x^3}\right)$ $+5\left(\frac{x}{3}\right)\left(\frac{1}{x^4}\right)+\frac{1}{x^5}$

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

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,

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

${=}^6{C}_0(x{)}^6{+}^6{C}_1(x{)}^5\left(\frac{1}{x}\right){+}^6{C}_2(x{)}^4{\left(\frac{1}{x}\right)}^2{+}^6{C}_3(x{)}^3{\left(\frac{1}{x}\right)}^3+$ ${}^6{C}_4(x{)}^2{\left(\frac{1}{x}\right)}^4{+}^6{C}_5(x){\left(\frac{1}{x}\right)}^5{+}^6{C}_6{\left(\frac{1}{x}\right)}^6$

$=x^6+6(x^5)\left(\frac{1}{x}\right)+15(x^4)\left(\frac{1}{x^2}\right)+20(8x^3)\left(\frac{1}{x^3}\right)$ $+15(x^2)\left(\frac{1}{x^4}\right)+6(x)\left(\frac{1}{x^5}\right)+\frac{1}{x^6}$

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

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

$\Rightarrow (102{)}^5$

$\Rightarrow (100+2{)}^5$

${=}^5{C}_0(100{)}^5{+}^5{C}_1(100{)}^4(2){+}^5{C}_2(100{)}^3(2{)}^2$ ${+}^5{C}_3(100{)}^2(2{)}^3{+}^5{C}_4(100{)}^1(2{)}^4{+}^5{C}_5(2{)}^5$

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

$=11040808032$

Question:8 Using binomial theorem, evaluate the following:

$(101{)}^4$

Answer:

As we can write 101 in the form 100 +1

$\Rightarrow (101{)}^4$

$\Rightarrow (100+1{)}^4$

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

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

$=104060401$

Question:9 Using binomial theorem, evaluate the following: $(99{)}^5$

Answer:

As we can write 99 in the form 100-1

$\Rightarrow (99{)}^5$

$\Rightarrow (100-1{)}^5$

${=}^5{C}_0(100{)}^5{-}^5{C}_1(100{)}^4(1){+}^5{C}_2(100{)}^3(1{)}^2$ ${-}^5{C}_3(100{)}^2(1{)}^3{+}^5{C}_4(100{)}^1(1{)}^4{-}^5{C}_5(1{)}^5$

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

$=9509900499$

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,

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

${=}^{10000}{C}_0{+}^{10000}{C}_1(1.1)$+Other positive terms

$=1+10000\times 1.1+$ Other positive term

$>1000$

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

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

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

From Here,

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

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

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

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 ,

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

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

From Here,

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

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

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

Now, Using this, we get

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

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

Question:13 Show that $9^{n+1}-8n-9$ is divisible by 64, whenevern 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,

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

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

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

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,

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

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

Hence Proved.

Question:1 Ifa andb 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,

$a=a-b+b$

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

${=}^n{C}_0(a-b{)}^n{+}^n{C}_1(a-b{)}^{n-1}b+.......{.}^n{C}_n{b}^n$

$=(a-b{)}^n{+}^n{C}_1(a-b{)}^{n-1}b+.......{.}^n{C}_{n-1}(a-b)b^{n-1}+b^n$ $\Rightarrow a^n-b^n=(a-b)[(a-b{)}^{n-1}{+}^n{C}_2(a-b{)}^{n-2}+........{+}^n{C}_{n-1}{b}^{n-1}]$

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

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

Question:2 Evaluate ${(\sqrt{3}+\sqrt{2})}^6-{(\sqrt{3}-\sqrt{2})}^6$.

Answer:

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

So,

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

$\Rightarrow (a+b{)}^6=a^6+6a^{5}b+15a^4{b}^2+20a^3{b}^3+15a^2{b}^4+6a{b}^5+b^6$

And

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

$\Rightarrow (a+b{)}^6=a^6-6a^{5}b+15a^4{b}^2-20a^3{b}^3+15a^2{b}^4-6a{b}^5+b^6$

Now,

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

$\Rightarrow (a+b{)}^6-(a-b{)}^6=2[6a^{5}b+20a^3{b}^3+6a{b}^5]$

Now, Putting $a=\sqrt{3}andb=\sqrt{2},$ we get

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

$\Rightarrow {(\sqrt{3}+\sqrt{2})}^6-{(\sqrt{3}-\sqrt{2})}^6=2[54\sqrt{6}+120\sqrt{6}+24\sqrt{6}]$

$\Rightarrow {(\sqrt{3}+\sqrt{2})}^6-{(\sqrt{3}-\sqrt{2})}^6=2\times 198\sqrt{6}$

$\Rightarrow {(\sqrt{3}+\sqrt{2})}^6-{(\sqrt{3}-\sqrt{2})}^6=396\sqrt{6}$

Question:3 Find the value of ${(a^2+\sqrt{a^2-1})}^4+{(a^2-\sqrt{a^2-1})}^4$

Answer:

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

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

$\Rightarrow (x+y{)}^4=x^4+4x^{3}y+6x^2{y}^2+4x{y}^3+y^4$

And

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

$\Rightarrow (x-y{)}^4=x^4-4x^{3}y+6x^2{y}^2-4x{y}^3+y^4$

Now,

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

$\Rightarrow (x+y{)}^4-(x-y{)}^4=2(x^4+6x^2{y}^2+y^4)$

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

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

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

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

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

Question:4 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,

$(0.99{)}^5=(1-0.001{)}^5{=}^5{C}_0(1{)}^5{-}^5{C}_1(1{)}^4(0.01){+}^5{C}_2(1{)}^3(0.01{)}^2$ $+othernegligibleterms$

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

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

$\Rightarrow (0.99{)}^5=0.951$

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

Question:5 Expand using Binomial Theorem ${(1+\frac{x}{2}-\frac{2}{x})}^4,x\ne 0$

Answer:

Given the expression,

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

The binomial expansion of this expression is

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

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

Now Applying the Binomial Theorem again,

${(1+\frac{x}{2})}^4{=}^4{C}_0(1{)}^4{+}^4{C}_1(1{)}^3\left(\frac{x}{2}\right){+}^4{C}_2(1{)}^2{\left(\frac{x}{2}\right)}^2{+}^4{C}_3(1){\left(\frac{x}{2}\right)}^3$ ${+}^4{C}_4{\left(\frac{x}{2}\right)}^4$

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

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

And

${(1+\frac{x}{2})}^3{=}^3{C}_0(1{)}^3{+}^3{C}_1(1{)}^2\left(\frac{x}{2}\right){+}^3{C}_2(1){\left(\frac{x}{2}\right)}^2{+}^3{C}_3{\left(\frac{x}{2}\right)}^3$

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