Binomial Theorem

Class 11 NCERT

NCERT

1   Expand the expression $(1-2x)^5$

Solution :

By using Binomial Theorem, the expression $(1-2x)^5$ can be expanded as $(1-2x)^5$$\\$ ${}^5C_{0}(1)^5 -{}^5C_1(1)^4(2x)+{}^5C_2(1)^3(2x)^2\\ -{}^5C_3(1)^2(2x)^3+{}^5C_4(1)^1(2x)^4-{}^5C_5(2x)^5\\ =1-5(2x)+10(4x)^2-10(8x^3)+5(16x^4)-(32x^5)\\ =1-10x+40x^2-80x^3+80x^4-32x^5$

2   Expand the expression $(\dfrac{2}{x}-\dfrac{x}{2})^5$

Solution :

By using Binomial Theorem, the expression $(\dfrac{2}{x}-\dfrac{x}{2})^5$can be expanded as$\\$ $(\dfrac{2}{x}-\dfrac{x}{2})^5\\ ={}^5C_0(\dfrac{2}{x})^5-{}^5C_1(\dfrac{2}{x})^4(\dfrac{x}{4})\\ +{}^5C_2(\dfrac{2}{x})^3(\dfrac{x}{2})^2\\ -{}^5C_3(\dfrac{2}{x})^2(\dfrac{x}{2})^3\\ +{}^5C_4(\dfrac{2}{x})(\dfrac{x}{2})^4-{}^5C_5(\dfrac{x}{2})^5$$\\$ $=\dfrac{32}{x^3}-5(\dfrac{16}{x^4})(\dfrac{x}{2})\\ +10(\dfrac{8}{x^3})(\dfrac{x^2}{4})\\ -10(\dfrac{4}{x^2})(\dfrac{x^3}{8})\\ +5(\dfrac{2}{x})(\dfrac{x^2}{16})-\dfrac{x^5}{32}$$\\$ $=\dfrac{32}{x^5}-\dfrac{40}{x^3}+\dfrac{20}{x}\\ -5x+\dfrac{5}{8}x^3-\dfrac{x^5}{32}$

3   Expand the expression $(2x-3)^6$

Solution :

By using Binomial Theorem, the expression $(2x-3)^6$ can be expanded as $(2x-3)^6\\ ={}^6C_0(2x)^6-{}^6C_1(2x)^5(3)+{}^6C_2(2x)^4(3)^2\\ -{}^6C_3(2x)^3(3)^3+{}^6C_4(2x)^2(3)^4\\ -{}^6C_5(2x)(3)^5+{}^6C_6(3)^6\\ =64x^6-6(32x^5)(3)+15(16x^4)(9)-20(8x^3)(27)\\ +15(4x^2)(81)-6(2x)(243)+729\\ =64x^6-576x^5+2160x^4-4320x^3+4860x^2-2916x+729$

4   Expand the expression $(\dfrac{x}{3}+\dfrac{1}{x})^5$

Solution :

By using Binomial Theorem, the expression $(\dfrac{x}{3}+\dfrac{1}{x})^5$ can be expanded as $\\$ $(\dfrac{x}{3}+\dfrac{1}{x})^5\\ ={}^5C_0(\dfrac{x}{3})^5+{}^5C_1(\dfrac{x}{3})^4(\dfrac{1}{x})\\ +{}^5C_2(\dfrac{x}{3})^3(\dfrac{1}{x})^2+{}^5C_3(\dfrac{x}{3})^2(\dfrac{1}{x})^3\\ +{}^5C_4(\dfrac{x}{3})(\dfrac{1}{x})^4+{}^5C_5(\dfrac{1}{x})^5\\ =\dfrac{x^5}{243}+5(\dfrac{x^4}{81})(\dfrac{1}{x})+10(\dfrac{x^3}{27})(\dfrac{1}{x^2})\\ +10(\dfrac{x^2}{9})(\dfrac{1}{x^3})+5(\dfrac{x}{3})(\dfrac{1}{x^4})+\dfrac{1}{x^5}\\ =\dfrac{x^5}{243}+\dfrac{5x^3}{81}\\ +\dfrac{10x}{9x}+\dfrac{5}{3x^3}+\dfrac{1}{x^5}$

5   Expand $(x+\dfrac{1}{x})^6$

Solution :

By using Binomial Theorem, the expression $(x+\dfrac{1}{x})^6$ can be expanded as$\\$ $(x+\dfrac{1}{x})^6\\ ={}^6C_0(x)^6+{}^6C_1(x)^5(\dfrac{1}{x})+\\ {}^6C_2(x)^4(\dfrac{1}{x})^2+{}^6C_3(x)^3(\dfrac{1}{x})^3\\ +{}^6C_4(x)^2(\dfrac{1}{x})^4+{}^6C_5(x)(\dfrac{1}{x})^5+{}^6C_6(\dfrac{1}{x})^6\\ =x^6+6(x)^5(\dfrac{1}{x})+15(x)^4(\dfrac{1}{x^2})+\\ 20(x)^3(\dfrac{1}{x^3})+15(x)^2(\dfrac{1}{x^4})6(x)(\dfrac{1}{x^5})+\dfrac{1}{x^6}\\ =x^6+6x^4+15x^2+20+\dfrac{15}{x^2}+\dfrac{6}{x^4}+\dfrac{1}{x^6}$

6   Using Binomial Theorem, evaluate $(96)^3$

Solution :

$96$ can be expressed as the sum or difference of two numbers whose powers are easier to calculate and then, binomial theorem can be applied.$\\$ It can be written that, $96 = 100 - 4$$\\$ $\therefore (96)^3=(100-4)^3\\ =^3C_0(100)^3-^3C_1(100)^2(4)+^3C_2(100)(4)^2-^3C_3(4)^3\\ =(100)^3-3(100)^2(4)+3(100)(4)^2-(4)^3\\ =1000000-120000+4800-64\\ =884736$

7   Using Binomial Theorem, evaluate $(102)^5$

Solution :

$102$ can be expressed as the sum or difference of two numbers whose powers are easier to calculate and then, binomial theorem can be applied.$\\$ It can be written that, $102 = 100 - 2$$\\$ $\therefore (102)^5=(100+2)^5\\ =^5C_0(100)^5+^5C_1(100)^4(2)+^5C_2(100)^3(2)^2+^5C_3(100)^2(2)^3+^5C_4(100)(2)^4+^5C_5(2)^5\\ =10000000000+1000000000+40000000+800000+8000+32\\ =11040808032$

8   Using Binomial Theorem, evaluate $( 101 )^ 4$

Solution :

$101$ can be expressed as the sum or difference of two numbers whose powers are easier to calculate and then, binomial theorem can be applied.$\\$ It can be written that, $101 = 100 + 1$$\\$ $\therefore (101)^2=(100+1)^4\\ =^4C_0(100)^4+^4C_1(100)^3(1)+^4C_2(100)^2(1)^2+^4C_3(100)(1)^3+^4C_4(1)^4\\ =(100)^4+4(100)^3+6(100)^2+4(100)+(1)^4\\ =100000000+4000000+60000+400+1\\ =104060401$

9   Using Binomial Theorem, evaluate $(99 ) ^5$

Solution :

$99$ can be written as the sum or difference of two numbers whose powers are easier to calculate and then, binomial theorem can be applied.$\\$ It can be written that,$ 99 = 100 - 1$$\\$ $\therefore (99)^5=(100-1)^5\\ =^5C_0(100)^5-^5C_1(100)^4(1)+^5C_2(100)^3(1)^2-^5C_3(100)^2(1)^3+^5C_4(100)(1)^4-^5C_5(1)^5\\ =(100)^5-5(100)^4+10(100)^3-10(100)^2+5(100)-1\\ =10000000000-500000000+10000000-100000+500-1\\ =10010000500-500100001\\ =9509900499$

10   Using Binomial Theorem, indicate which number is larger $(1.1)^{10000} \text{or} 1000$

Solution :

By splitting $1.1$ and then applying Binomial Theorem, the first few terms of $(1.1)^{10000}$ be obtained as $\\$ $(1.1)^{10000}=(1+0.1)^{10000}\\ =^{10000}C_0+^{10000}C_1(1.1)+\text{Other positive terms}\\ =1+10000*1.1+\text{Other positive terms}\\ =1+11000+\text{Other positive terms}\\ > 1000$$\\$ Hence,$(1.1)^{10000} > 1000.$

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

Solution :

Using Binomial Theorem, the expressions, $( a + b)^4$ and $( a - b )^4,$ can be expanded as $\\$ $( a + b)^4 =^4C_0a^4+^4C_1a^3b+^4C_2a^2b^2+^4C_3ab^3+^4C_4b^4\\ (a-b)^4=^4C_0a^4-^4C_1a^3b+^4C_2a^2b^2-^4C_3ab^3+^4C_4b^4\\ \therefore (a+b)^4-(a-b)^4=^4C_0a^4+^4C_1a^3b+^4C_2a^2b^2+^4C_3ab^3+^4C_4b^4-[^4C_0a^4-^4C_1a^3b+^4C_2a^2b^2-^4C_3ab^3+^4C_4b^4]\\ =2(^4C_1a^3b+^4C_3ab^3)=2(4a^3b+4ab^3)\\ =8ab(a^2+b^2)$ $\\$ By putting $a =\sqrt{ 3}$ and $b =\sqrt{ 2}$ , we obtain$\\$ $(\sqrt{3}+\sqrt{2})^4-(\sqrt{3}-\sqrt{2})^4=8(\sqrt{3})(\sqrt{2})\{(\sqrt{3})^2+(\sqrt{2})^2\}\\ =8(\sqrt{6})\{3+2\}=40\sqrt{6}$

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

Solution :

Using Binomial Theorem, the expression,$(x+1)^6$ and $(x-1)^6$ , can be expanded as $\\$ $(x+1)^6=^6C_0x^6+^6C_1x^5+^6C_2x^4+^6C_3x^3+^6C_4x^2+^6C_5x+^6C_6\\ (x-1)^6=^6C_0x^6+^6C_1x^5+^6C_2x^4+^6C_3x^3+^6C_4x^2+^6C_5x+^6C_6\\ \therefore (x+1)^6+(x-1)^6 =2[^6C_0x^6+^6C_2x^4+^6C_4x^2+^6C_6]\\ =2[x^6+15x^4+15x^2+1]$ $\\$ By putting $ x=\sqrt{2}$ we obtain$ \\$ $(\sqrt{2}+1)^6+(\sqrt{2}-1)^6=2[(\sqrt{2})^6+15(\sqrt{2})^4+15(\sqrt{2})^2+1]\\ =2(8+15*4+15*2+1)\\ =2(8+60+30+1)\\ =2(99)=198$

13   Show that $9^{n+1}-8n-9$ is divisible by 64, whenever n is a positive integer.

Solution :

In order to show that $9^{n+1}-8n-9$ is divisible by 64, it has to be prove that,$9^{n+1}-8n-9=64k,$ where k is some natural number$\\$ By Binomial Theorem,$\\$ $(1+a)^m=^mC_0+^mC_1a+^mC_2a^2+.....+^mC_ma^m$ $\\$ For $a = 8$ and $m = n + 1$ , we obtain $\\$ $(1+8)^{n+1}=^{n+1}C_0+^{n+1}C_1(8)+^{n+1}C_2(8)^2+.....+^{n+1}C_{n+1}(8)^{n+1}\\ \implies 9^{n+1}=1+(n+1)(8)+8^2[^{n+1}C_2+^{n+1}C_3*8+...+^{n+1}C_{n+1}(8)^{n-1}]\\ \implies 9^{n+1}=9+8n+64[^{n+1}C_2+^{n+1}C_3*8+....+^{n+1}C_{n+1}(8)^{n-1}]\\ \implies 9^{n+1}-8n-9$ is divisible by 64, whenever n is a positive integer.

14   Prove that $ \displaystyle\sum_{r=0}^{n} 3^r \ \ $$ \ \ ^nC_r=4^n$

Solution :

By Binomial Theorem, $\\$ $ \displaystyle\sum_{r=0}^{n} 3^r \ \ $$ \ \ ^nC_r a^{n-r}b^r=(a+b)^n$$\\$ By putting $ b = 3$ and $a = 1$ in the above equation, we obtain $ \displaystyle\sum_{r=0}^{n} 3^r \ \ $$ \ \ ^nC_r 1^{n-r}3^r=(1+3)^n$$\\$ $\implies \displaystyle\sum_{r=0}^{n} 3^r \ \ $$ \ \ ^nC_r =4^n$ $\\$ Hence proved.

15   Find the coefficient of $x^ 5$ in $( x + 3 )^ 8$

Solution :

It is known that $(r+1)^{th}$ term $(T_{r+1}),$ in the binomial expansion of $(a+b)^n$ is given by $ \\$$T_{r+1}=^nC_r a^{n-r} b^r$ $\\$ Assuming that $x^ 5$ occurs in the $( r + 1 )^{th}$ term of the expansion $(x+3)^8$ , we obtain$\\$ $T_{r+1}=^nC_r x^{8-r} 3^r$ $\\$ Comparing the indices of x in $x^ 5$ in $T _{r + 1}$ ,$\\$ We obtain $r = 3$ $\\$ Thus, the coefficient of $x^5$ is $ ^8C_3 (3)^3=\dfrac{8!}{3!5!}*3^3=\dfrac{8.7.6.5!}{3.2.5!}.3^3=1512$

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

Solution :

It is known that $(r+1)^{th}$ term,$(T_{r+1}),$ in the binomial expansion of $(a+b)^n$ is given by $T_{r+1}=^nC_r a^{n-r} b^r$ $\\$ Assuming that $a ^5 b ^7 $ occurs in the $(r+1)^{th}$ term of the expansion $(a-2b)^{12},$ we obtain $ T{r+1}=^{12}C_r(a)^{12-r} (-2b)^r=^{12}C_r(-2)^r(a)^{12-r}(b)^r$ $\\$ Comparing the indices of a and b in $ a^ 5 b ^7$ in $T_{r+1},$ $\\$ We obtain $r = 7$$\\$ Thus, the coefficient of $a ^5 b ^7$ is $\\$ $^{12}C_7(-2)^7=\dfrac{12!}{7!5!}.2^7=\dfrac{12.11.10.9.8.7!}{5.4.3.2.7!}.(-2)^7=-(792)(128)=-101376.$

17   Write the general term in the expansion of $(x^2-y)^6$

Solution :

It is known that the general term $T_{r+1}$which is the $(r+1)^{th}$ term in the binomial expansion of $(a+b)^n$ is given by $ T_{r+1}=^nC_r a^{n-r} b^r.$ $\\$ Thus, the general term in the expansion of $(x^2-y^6)$ is $\\$ $T_{r+1}=^6C_r(x^2)^{6-r}(-y)^r=(-1)^r ^6C_r.x^{12-2r}.y^r$

It is known that the general term $T_{r+1}$which is the $(r+1)^{th}$ term in the binomial expansion of $(a+b)^n$ is given by $ T_{r+1}=^nC_r a^{n-r} b^r.$ $\\$ Thus, the general term in the expansion of $(x^2-y^6)$ is $\\$ $T_{r+1}=^6C_r(x^2) $$^{6-r}(-y) ^r=(-1)^r $$ ^6C_r.x^{12-2r}.y^r$

18   Write the general term in the expansion of $(x^2-yx)^{12},x\neq 0$

Solution :

It is known that the general term $T_{r+1}$ which is the $(r+1)^{th}$ term} in the binomial expansion of $(a+b)^n$ is given by $ T_{r+1}=^nC_r a^{n-r} b^r.$ $\\$ Thus, the general term in the expansion of $(x^2-yx)^{12}$ is $T_{r+1}=^{12}C_r(x^2)$ $^{12-r} (-yx)^r=(-1)^r $ $^{12}C_r. x^{24-2r}.y^{r}=(-1)^r$ $^{12}C_r.x^{24-r}.y^r$

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

Solution :

It is known $(r+1)^{th}$term, $T_{r+1},$ in the binomial expansion of $(a+b)^n$ is given by $T_{r+1}=^nC_r a^{n-r}b^r.$ $\\$ Thus, the $4^{ th}$ term in the expansion of $(x^2-2y)^{12}$ is $\\$ $T_4=T_{3+1}=^{12}C_3(x)^{12-3}(-2y)^3=\\(-1)^3.\dfrac{12!}{3!9!}.x^9.(2)^3.y^3=-\dfrac{12.11.10}{3.2}.(2)^3x^9y^3=-1760x^9y^3$

20   Find the ${13^{th}}$ term in the expansion of$(9x-\dfrac{1}{3\sqrt{x}})^{18},x\neq 0$

Solution :

It is known $(r+1)^{th}$ term, $ T_{r+1}$, in the binomial expansion of $(a+b)^n$ is given by $T_{r+1}=n^C_r a^{n-r}b^r$ $\\$ Thus, the $13^{th}$ term in the expansion of $(9x-\dfrac{1}{3\sqrt{x}})^{18}$ is $\\$ $ T_{13}=T_{12+1}=^{18}C_{12}(9x)^{18-12}(-\dfrac{1}{3\sqrt{x}})^{12}\\ =(-1)^{12}\dfrac{18!}{12!6!}(9)^6(x)^6(\dfrac{1}{3})^{12}(\dfrac{1}{\sqrt{x}})^{12}\\ =\dfrac{18.17.16.15.14.13.12!}{12!.6.5.4.3.2}.x^6(\dfrac{1}{x^6}).3^{12}(\dfrac{1}{3^{12}})\quad [9^6=(3^2)^6=3^{12}]\\ =18564$

21   Find the middle terms in the expansions of $(3-\dfrac{x^3}{6})^7$

Solution :

It is known that in the expansion of $(a+b)^n$, in n is odd, then there are two middle terms,$\\$ Namely $(\dfrac{n+1}{2})^{th}$ term and $(\dfrac{n+1}{2}+1)^{th}$ term.$\\$ Therefore, the middle terms in the expansion $(3-\dfrac{x^3}{6})^7$ are $(\dfrac{7+1}{2})^{th}=4^{th}$ and $(\dfrac{7+1}{2}+1)^{th}=5^{th}$ $\\$ term $\\$ $T_4=T_{3+1}=^7C_3(3)^{7-3}(-\dfrac{x^3}{6})^3=(-1)^3\dfrac{7!}{3!4!}.3^4.\dfrac{x^9}{6^3}\\ =-\dfrac{7.6.5.4!}{3.2.4!}.3^4.\dfrac{1}{2^3.3^3}.x^9=-\dfrac{105}{8}x^9\\ T_5=T_{4+1}=^7C_4(3)^{7-4}(-\dfrac{x^3}{6})^4=(-1)^4\dfrac{7!}{4!3!}.3^3.\dfrac{x^{12}}{6^4}\\ =\dfrac{7.6.5.4!}{4!.3.2}.\dfrac{3^3}{2^4.3^4}.x^{12}=\dfrac{35}{48}x^{12}$ $\\$ Thus, the middle terms in the expansion of $(3-\dfrac{x^3}{6})^7$ are $ -\dfrac{105}{8}x^9$ and $\dfrac{35}{48}x^{12}.$

22   Find the middle terms in the expansion of $(\dfrac{x}{3}+9y)^{10}$

Solution :

It is known that in the expansion of $(a+b)^n,$ in n is even, then the middle term is $(\dfrac{n}{2}+1)^{th}$ term.$\\$ Therefore, the middle term in the expansion of $(\dfrac{x}{3}+9y)^{10}$ is $(\dfrac{10}{2}+1)^{th}=6^{th}\\ T_4=T_{5+1}=^{10}C_5 (\dfrac{x}{3})^{10-5}(9y)^5\\ =\dfrac{10!}{5!5!}.\dfrac{x^5}{3^5}.9^5.y^5\\ =\dfrac{10.9.8.7.6.5!}{5.4.3.2.5!}.\dfrac{1}{3^5}.3^{10}.x^5y^5 \quad [9^5=(3^2)^5=3^{10}]\\ =252*3^5.x^5.y^5=6123x^5y^5$ $\\$ Thus, the middle term in the expansion of $(\dfrac{x}{3}+9y)^{10}$ is $ 61236x^5y^5.$

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

Solution :

It is known that $(r+1)^{th}$ term, $(T_{r+1}),$ in the binomial expansion of $(a+b)^n$ is given by $\\$ $T_{r+1}=^nC_r a^{n-r} b^r$ $\\$ Assuming that $a^ m$ occurs in the $(r+1)^{th}$ term of the expansion $(1+a)^{m+n}$, we obtain $\\$ $T_{r+1}=^{m+n}C_r(1)^{m+n-r}(a)^r=^{m+n}C_r a^r$ $\\$ Comparing the indices of a in $a^m$ in $ T_{r+1},$ $\\$ We obtain $ r=m$ $\\$ Therefore, the coefficient of $ a^m$ is $\\$ $ ^{m+n}C_m=\dfrac{(m+n)!}{m!(m+n-m)!}=\dfrac{(m+n)!}{m!n!}.....(1)$ $\\$ Assuming that $a^n$ occurs in the $(k+1)^{th}$ term of the expansion $(1+a)^{m+n},$ we obtain $\\$ $T_{k+1}=^{m+n}C_k(1)^{m+n-k}(a)^k=^{m+n}C_k(a)^k$ $\\$ Comparing the indices of a in $a^n$ and in $T_{k+1,}$ $ \\$ we obtain $ \\$ $k=n$ $\\$ Therefore, the coefficient of $ a^n$ is $\\$ $^{m+n}C_n=\dfrac{(m+n)!}{n!(m+n-n)!}=\dfrac{(m+n)!}{n!m!}.....(2)$ $\\$ Thus, from (1) and (2), it can be observed that the coefficients of $ a^m$ and $ a^n $ in the expansion of $(1+a)^{m+n}$ are equal.

24   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.

Solution :

It is known that $(k+1)^{th}$ term , $(T_{k+1})$, in the binomial expansion of $(a+b)^n$ is given by $ \\$ $T_{k+1}=^nC_k a^{n-k} b^k. $ $\\$ Therefore, $(r-1)^th$ term in the expansion of $(x+1)^n$ is $\\$ $T_{r-1}=^nC_{r-2}(x)^{n-(r-2)}(1)^{(r-2)}=^nC_{r-2}x^{n-r+2}$ $\\$ $(r+1)$ term in the expansion of $(x+1)^n$ is $\\$ $T{r-1}=^nC_r(x)^{n-r}(1)^r=^nC_rx^{n-r}$ $\\$ $r^{th}$term in the expansion of $(x+1)^n $ is $\\$ $T_r=^nC_{r-1}(x)^{n-(r-1)}(1)^{(r-1)}=^nC_{r-1}x^{n-r+1}$ $\\$ Therefore, the coefficients of the $(r-1)^{th},r^{th}$ and $(r+1)^th$ terms in the expansion of $(x+1)^n $ $\\$ $^nC_{r-2},^nC_r-1,$ and $^nC_r$ are respectively. Since these coefficients are in the ratio $1:3:5,$ we obtain $\\$ $\dfrac{^nC_{r-2}}{^nC_{r-1}}=\dfrac{n!}{(r-2)!(n-r+2)!}*\dfrac{(r-1)!(n-r+1)!}{n!}=\\ \dfrac{(r-1)(r-2)!(n-r+1)!}{(r-2)!(n-r+2)!(n-r+1)!}\\ =\dfrac{r-1}{n-r+2}\\ \therefore \dfrac{r-1}{n-r+2}=\dfrac{1}{3}\\ \implies 3r-3=n-r+2\\ \implies n-4r+5=0....(1)\\ \dfrac{^nC_{r-1}}{^nC_r}=\dfrac{n!}{(r-1)!(n-r+1)}*\dfrac{!(n-r)!}{}n!\\ =\dfrac{r(r-1)!(n-r)!}{(r-1)!(n-r+1)(n-r)!}\\ =\dfrac{r}{n-r+1}\\ \therefore \dfrac{r}{n-r+1}=\dfrac{3}{5}\\ \implies 5r=3n-3r+3\\ \implies 3n-8r+3=0......(2)$ $\\$ Multiplying (1) by 3 and subtracting it from (2), we obtain $\\$ $4r-12=0\\ \implies r=3$ $\\$ Putting the value of r in (1) , we obtain n$\\$ $-12+5=0\\ \implies n=7$$\\$ Thus, $n=7$ and $ r=3 $

25   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 $(1+x)^{2n-1}$

Solution :

It is known that $(r+1)^{th}$ term, $ (T_{r+1})$ in the binomial expansion of $(a+b)^n$ is given by $T_{r+1}=^nC_r a^{n-r} b^r . $ $\\$ Assuming that $x^n$ occurs in the $(r+1)^{th}$ term of the expansion of $(1+x)^{2n}$,we obtain $\\$ $T_{r+1}=^{2n}C_r (1)^{2n-r} (x)^r=^{2n}C_r(x)^r$ $\\$ Comparing the indices of x in $x^n$ and in $T_{r+1},$ we obtain $r=n$ $\\$ Therefore, the coefficient of $x^n$ in the expansion of $(1+x)^{2n}$ is $\\$ $^{2n}C_n=\dfrac{(2n)!}{n!(2n-n)!}=\dfrac{(2n)!}{n!n!}=\dfrac{(2n)!}{(n!)^2}.....(1)$ $\\$ Assuming that $x^n$ occurs in the $(k+1)^{th}$ term of the expansion of $(1+x)^{2n-1}$,we obtain $\\$ $T_{k+1}=^{2n}C_k(1)^{2n-1-k}(x)^k$ $\\$ Comparing the indices of x in $x^n$ and in $ T_{k+1},$ we obtain $ k=n$ $\\$ Therefore, the coefficient of $x^ n$ in the expansion of $(1+x)^{2n-1}$ is $\\$ $^{2n-1}C_n=\dfrac{(2n-1)!}{n!(2n-1-n)!}=\dfrac{(2n-1)!}{n!(n-1)!}\\ =\dfrac{2n.(2n-1)!}{2n.n!(n-1)!}=\dfrac{(2n)!}{2.n!n!}=\dfrac{1}{2}[\dfrac{(2n)!}{(n!)^2}]......(2)$ $\\$ From (1) and (2) , it is observed that $\\$ $\dfrac{1}{2}(^{2n}C_n)=^{2n-1}C_n\\ \implies ^{2n}C_n=2(^{2n-1}C_n)$ $\\$ Therefore, the coefficient of $x^n$ expansion of $(1+x)^{2n}$ is twice the coefficient of $x^n$ in the expansion of $(1+x)^{2n-1}.$$\\$ Hence proved.

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

Solution :

It is known that $(r+1)^{th}$ term,$(T_{r+1}),$ in the binomial expansion of $(a+b)^n$ is given by $\\$ $T_{r+1}=^nC_r a^{n-r} b^r$ $\\$ Assuming that $x^2$ occurs in the $(r+1)^{th}$ term of the expansion of $(1+x)^m,$ we obtain $ \\$$T_{r+1}=^mC_r(1)^{m-r}(x)^r=^mC_r(x)^r$$\\$ Comparing the indices of x in $x^2$ and in $T_{r+1},$ we obtain $ r=2$ $\\$ Therefore, the coefficient of $x^2$ is $^mC_2$ $\\$ It is given that the coefficient of $x^2$ in the expansion$(1+x)^m$ is 6.$\\$ $\therefore ^mC_2=6\\ \implies \dfrac{m!}{2!(m-2)!}=6\\ \implies \dfrac{m(m-1)(m-2)!}{2*(m-2)!}=6\\ \implies m(m-1)=12\\ \implies m^2-m-12=0\\ \implies m^2-4m+3m-12 =0\\ \implies m(m-4)+3(m-4)=0\\ \implies (m-4)(m+3)=0\\ \implies (m-4)=0 or (m+3)=0\\ \implies m=4 \ or \ m=-3$ $\\$ Thus, the positive value of m, for which the coefficient of$x^2$ in the expansion $(1+x)^m$ is 6, is 4.

27   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.

Solution :

It is known that $(r+1)^{th}$ term, $(T_{r+1}),$ in the binomial expansion of $(a+b)^n $ is given by $\\$ $T_{r+1}=^nC_r a^{n-r} b^r.$ $\\$ The first three terms of the expansion are given as $ 729 $,$ 7290 $ and $ 30375 $ respectively.$\\$ Therefore, we obtain $\\$ $T_1=^C_0 a^{n-0}b^0=a^n=729.....(1)\\ T_2=^nC_1 a^{n-1} b^1=na^{n-1} b=7290.....(2)\\ T_2=^nC_1 a^{n-2} b^2=\dfrac{n(n-1)}{2}a^{n-2}b^2 =30375.....(3)$ $\\$ Dividing (2) by (1), we obtain $\\$ $\dfrac{na^{n-1}b}{a^n}=\dfrac{7290}{7290}\\ \implies \dfrac{(n-1)b}{2a}=\dfrac{30375}{7290}\\ \implies \dfrac{(n-1)b}{a}=\dfrac{30375*2}{7290} =\dfrac{25}{3}\\ \implies \dfrac{nb}{a}-\dfrac{b}{a}=\dfrac{25}{3}\\ \implies 10-\dfrac{b}{a}=\dfrac{25}{3} \quad [\text{Using (4)}] \\ \implies \dfrac{b}{a} =10-\dfrac{25}{3}=\dfrac{5}{3}.....(5)$ $\\$ From (4) and (5), we obtain $\\$ $n.\dfrac{5}{3}=10\\ \implies n=6 $ $\\$ Substituting $n = 6$ in equation (1), we obtain $a ^6$ $\\$ $=729\\ \implies a={^6\sqrt{729}}=3 $ $\\$ From (5),we obtain $\\$ $\dfrac{b}{3}=\dfrac{5}{3}\implies b=5 $ $\\$ Thus,$ a=3,b=5,$ and $ n=6.$

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

Solution :

It is known that $(r+1)^{th}$ tern $(T_{r+1}),$ in the binomial expansion of $(a+b)^n$ is given by $\\$ $T_{r+1}=^nC_r a^{n-r} b^r.$ $\\$ Assuming that $x^2$ occurs in the $(r+1)^{th}$ term in the expansion of $(3+ax)^9,$ we obtain $\\$ $T_{r+1}=^9C_r(3)^{9-r}(ax)^r=^9C_r(3)^{9-r} a^r x^r$ $\\$ Comparing the indices of $x$ in $x^2$ and in $T_{r+1},$ we obtain $\\$ $r=2 $ $\\$ Thus, the coefficient of $x^2$ is $\\$ $^9C_2(3)^{9-2} a^2=\dfrac{9!}{2!7!}(3)^7 a^2=36(3)^7 a^2$ $\\$ Assuming that $x^3$ occurs in the $(k+1)^{th}$ term in the expansion of $(3+ax)^9,$ we obtain $\\$ $T_{k+1}=^9C_k(3)^{9-k} (ax)^k=^9C_k(3)^{9-k} a^k x^k$ $\\$ Comparing the indices of $x$ in $x^3$ and in $T_{k+1},$ we obtain $ k=3$ $\\$ Thus, the coefficient of $x^3$ is $\\$ $^9C_3(3)^{9-3} a^3=\dfrac{9!}{3!6!}(3)^6 a^3=84(3)^6 a^3 $ $\\$ It is given that the coefficient of $x^2$ and $x^3$ are the same. $\\$ $84(3)^6 a^3=36(3)^7 a^2\\ \implies 84 a=36*3\\ \implies a=\dfrac{36*3}{84}=\dfrac{104}{84}\\ \implies a=\dfrac{9}{7}$ $\\$ Thus, the required value of a is $9/7.$

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

Solution :

Using Binomial Theorem, the expressions, $(1+2x)^6$ and $(1-x)^7,$ can be expanded as $\\$ $(1+2x)^6=^6C_0+^6C_1(2x)+^6C_2(2x)^2+^6C_4(2x)^4+^6C_5(2x)^5+^6C_6(2x)^6\\ =1+6(2x)+15(2x)^2+20(2x)^3+15(2x)^4+6(2x)^5+(2x)^6\\ 1+12x+60x^2+160x^3+240x^4+192x^5+64x^6\\ (1-x)^7=^7C_0-^7C_1(x)+^7C_2(x)^2-^7C_3(x)^3+^7C_4(x)^4-^7C_5(x)^5+^7C_6(x)^6-^7C_7(x)^6\\ =1-7x+21x^2-35x^3+35x^4-21x^5+7x^6-x^7\\ \therefore (1+2x)^6(1-x)^7\\ =(1+12x+60x^2+160x^3+240x^4+192 x^5+64x^6)(1-7x+21x^2-35x^3+35x^4-21x^5+7x^6-x^7)$ $\\$ The complete multiplication of the two brackets is not required to be carried out. Only those terms, which involve $x^ 5$ , are required.$\\$ The terms containing $x ^5$ are$\\$ $1(-21x^5)+(12x)(35x^4)+(60x^2)(-35x^3)+(160x^3)(21x^2)+(240x^4)(-7x)+(192 x^5)(1)\\ =171^5$ $\\$ Thus, the coefficient of $x^ 5$ in the given product is $171$.

30   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 $a^ n =(a - b + b )$ and expand]

Solution :

In order to prove that $(a-b)$ is a factor of $(a^n-b^n)$ it has to be proved that $a^n-b^n=k(a-b),$ where $ k $ is some natural number$\\$ It can be written that, $a=a-b+b$ $\\$ $\therefore a^n=(a-b+b)^n=[(a-b)+b]^n\\ =^nC_0(a-b)^n+^nC_1(a-b)^{n-1} b+...+^nC_{n-1}(a-b)b^{n-1}+^nC_n b^n\\ =(a-b)^n+^nC_1(a-b)^{n-1}b+....+^nC_{n-1}(a-b)b^{n-1}+b^n\\ \implies a^n-b^n=(a-b)[(a-b)^{n-1}+^nC_1(a-b)^{n-2} b+.....+^nC_{n-1} b^{n-1}]\\ \implies a^n-b^n=k(a-b)$ $\\$ Where,$k=[(a-b)^{n-1}+^nC_1(a-b)^{n-2}b+....+^nC_{n-1} b^{n-1}]$ is a natural number $\\$ This shows that $(a-b)$ is a factor of $(a^n-b^n),$ where $n$ is a positive integer.

31   Evaluate $(\sqrt{3}+\sqrt{2})^6-(\sqrt{3}-\sqrt{2})^6$

Solution :

Firstly, the expression $(a+b)^6-(a-b)^6$ is simplified by using Binomial Theorem. This can be done as $\\$ $(a+b)^6=^6C_0a^6+^6C_1a^5b+^6C_2a^4b^2+^6C_3a^3b^3+^6C_4 a^2b^4+^6C_5 a^1b^5+^6C_6b^6\\ =a^6+6a^5b+15a^4b^2+20a^3b^3+15a^2b^4+6ab^5+b^6\\ (a-b)^6=^6C_0a^6-^6C_1a^5b+^6C_2a^4b^2-^6C_3a^3b^3+^6C_4 a^2b^4-^6C_5 a^1b^5+^6C_6b^6\\ =a^6-6a^5b+15a^4b^2-20a^3b^3+15a^2b^4-6ab^5+b^6\\ \therefore (a+b)^6-(a-b)^6=2[6a^5b+20a^3b^3+6ab^5]$ $\\$ Putting $a=\sqrt{3}$ and $b=\sqrt{2},$ we obtain$\\$ $(\sqrt{3}+\sqrt{2})^6-(\sqrt{3}-\sqrt{2})^6=2[6(\sqrt{3})^5(\sqrt{2})+20(\sqrt{3})^3(\sqrt{2})^3+6(\sqrt{3})(\sqrt{2})^5]\\ =2[54\sqrt{6}+120\sqrt{6}+24\sqrt{6}]\\ =2*198\sqrt{6}\\ =396\sqrt{6}$

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

Solution :

Firstly, the expression $(x+y)^4+(x-y)^4$ is simplified by using Binomial Theorem. $\\$ This can be done as $\\$ $(x+y)^4=^4C_0x^4+^4C_1x^3y+^4C_2x^2y^2+^4C_3xy^3+^4C_4y^4\\ =x^4+4x^3y+6x^2y^2+4xy^3+y^4\\ (x-y)^4=^4C_0x^4-^4C_1x^3y+^3C_2x^2y^2-^4C_3xy^3+^4C_4y^4\\ =x^4-4x^3y+6x^2y^2-4xy^3+y^4\\ \therefore (x+y)^4+(x-y)^4=2(x^4+6x^2y^2+y^4)$ $\\$ Putting $ x=a^2$ and $y=\sqrt{a^2-1}$ we obtain$\\$ $(a^2+\sqrt{a^2-1})^4+(a^2-\sqrt{a^2-1})^4=\\ 2[(a^2)^4+6(a^2)^2 (\sqrt{a^2-1})^2+(\sqrt{a^2-1})^4]\\ =2[a^8+6a^4(a^2-1)+(a^2-1)^2]\\ =2[a^8+6a^6-6a^4+a^4-2a^2+1]\\ =2[a^8+6a^6-5a^4-2a^2+1]\\ =2a^8+12a^6-10a^4-4a^2+2$

33   Find an approximation of $(0.99)^5$using the first three terms of its expansion.

Solution :

$0.99 = 1 - 0.01\\ \therefore ( 0.99 )^5 =( 1 - 0.01 )^5\\ = ^5 C_0 (1 )^5-^5 C_ 1 ( 1 )^4( 0.01 )+^ 5 C_ 2 ( 1)^3( 0.01 )^2 \quad [Approximately] \\ = 1 - 5 ( 0.01 )+ 10 ( 0.01 )^2\\ = 1 - 0.05 + 0.001\\ = 1.001 - 0.05\\ = 0.951 $ $\\$ Thus, the value of $( 0.99 )^5$ is approximately $0.951$.

34   Find $n$, if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of $(^\sqrt{2}+\dfrac{1}{4\sqrt{3}})^n$ is $\sqrt{6}:1.$

Solution :

In the expansion, $(a+b)^n=^nC_0a^n+^C_1a^{n-1}b+^nC_2a^{n-2}b^2+......+^nC_{n-1}ab^{n-1}+^nC_nb^n$ $\\$ Fifth term from the beginning $=^nC_4a^{n-4}b^4$ $\\$ Fifth term from the end $=^nC_4a^4b^{n-4}$ $\\$ Therefore, it is evident that in the expansion of $(^4\sqrt{2}+\dfrac{1}{^4\sqrt{3}})^n$ are fifth term from the beginning is $^nC_4(^4\sqrt{2})^{n-4}(\dfrac{1}{^4\sqrt{3}})^4$ and the fifth term from the end is $^nC_{n-4}(^\sqrt{2})^2(\dfrac{1}{^4\sqrt{3}})^{n-4}$ $\\$ $^nC_4(^4\sqrt{2})^{n-4}(\dfrac{1}{^4\sqrt{3}})^4=^nC_4\dfrac{(^4\sqrt{2})^n}{(^4\sqrt{2})^4}.\dfrac{1}{3}=\dfrac{n!}{6.4!(n-4)!}(^4\sqrt{2})^n......(1)\\ ^nC_{n-4}(^4\sqrt{2})^4(\dfrac{1}{^4\sqrt{3}})^{n-4}=\\ ^nC_{n-4}\dfrac{(^4\sqrt{3})^4}{(^4\sqrt{3})^n} =^nC_{n-4}.2.\dfrac{3}{(^4\sqrt{3})^n}=\dfrac{6n!}{(n-4)!4!}.\dfrac{1}{(^4\sqrt{3})^n}.....(2)$ $\\$

It is given that the ratio of the fifth term from the beginning to the fifth term from the end is $\sqrt{6} :1$ . Therefore, from (1) and (2), we obtain $\\$ $\dfrac{n!}{6.4!(n-4)!}(^4\sqrt{2})^n:\dfrac{6n!}{(n-4)!4!}.\dfrac{1}{(^4\sqrt{3})^n}=\sqrt{6}:1\\ \implies \dfrac{(^4\sqrt{2})^n}{6}:\dfrac{6}{(^4\sqrt{3})^n}=\sqrt{6}:1\\ \implies \dfrac{(^4\sqrt{2})^n}{6}*\dfrac{(^4\sqrt{3})^n}{6}=\sqrt{6}\\ \implies (^4\sqrt{6})^n =36\sqrt{6}\\ \implies 6^{n/4}=\dfrac{5}{2}\\ \implies n=4*\dfrac{5}{2}=10$ $\\$ Thus, the value of $ n $ is $10.$

35   Expand using Binomial Theorem $(1+\dfrac{x}{2}-\dfrac{2}{x})^4,x\neq 0$

Solution :

$(1+\dfrac{x}{2}-\dfrac{2}{x})^4\\ =^nC_0(1+\dfrac{x}{2})^4-^nC_1(1+\dfrac{x}{2})^3(\dfrac{2}{x})+\\ ^nC_2(1+\dfrac{x}{2})^2(\dfrac{2}{x})^2-^nC_3(1+\dfrac{x}{2})(\dfrac{2}{x})^3 +^nC_4(\dfrac{2}{x})^4\\ =(1+\dfrac{x}{2})^4-4(1+\dfrac{x}{2})^3(\dfrac{2}{x})+\\ 6(1+x+\dfrac{x^2}{2})(\dfrac{4}{x^2})-4(1+\dfrac{x}{2})(\dfrac{8}{x^3 })+(\dfrac{16}{x^4})\\ =(1+\dfrac{x}{2})^4-\dfrac{8}{x}(1+\dfrac{x}{2})^3+\dfrac{24}{x^2}+\dfrac{24}{x}+6-\dfrac{32}{x^3}-\dfrac{16}{x^2}+\dfrac{16}{x^4}\\ =(1+\dfrac{x}{2})^4-\dfrac{8}{x}(1+\dfrac{x}{2})^3+\dfrac{8}{x^2}+\dfrac{24}{x}+6-\dfrac{32}{x^3}+\dfrac{16}{x^4}.....(1)$ $\\$ Again by using Binomial Theorem, we obtain $\\$ $(1+\dfrac{x}{2})^4=^4C_0(1)^4+^4C_1(1)^3(\dfrac{x}{2})+^4C_2(1)^2(\dfrac{x}{2})^2+^4C_3(1)^3(\dfrac{x}{2})^4\\ =1+4*\dfrac{x}{2}+6*\dfrac{x^2}{4}+4*\dfrac{x^3}{8}+\dfrac{x^4}{16}\\ =1+2x+\dfrac{3x^2}{2}+\dfrac{x^3}{2}+\dfrac{x^4}{16}.....(2)\\ (1+\dfrac{x}{2})^3=^3C_0(1)^3+^3C_1(1)^2(\dfrac{x}{2})+^3C_2(1)(\dfrac{x}{2})+^3C_3(\dfrac{x}{2})^3\\ =1+\dfrac{3x}{2}+\dfrac{3x^2}{4}+\dfrac{x^3}{8}......(3)$ $\\$ From (1), (2) and (3), we obtain $\\$ $[(1+\dfrac{x}{2})-\dfrac{2}{x}]^4\\ =1+2x+\dfrac{3x^2}{2}+\dfrac{x^3}{2}+\dfrac{x^4}{16}-\dfrac{8}{x}-12-6x-x^2+\dfrac{8}{x^2}+\dfrac{24}{x}+6-\dfrac{32}{x^3}+\dfrac{16}{x^4}\\ =\dfrac{16}{x}+\dfrac{8}{x^2}-\dfrac{32}{x^3}+\dfrac{16}{x^4}-4x+\dfrac{x^2}{2}+\dfrac{x^3}{2}+\dfrac{x^4}{16}-5$

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

Solution :

Using Binomial Theorem, the given expression $(3x^2-2ax+3a^2)^3$ can be expanded as $[(3x^2-2ax)+3a^2]^3$ $\\$ $=^3C_0(3x^2-2ax)^3+^3C_1(3x^2-2ax)^2(3a^2)+^3C_2(3x^2-2ax)(3a^2)^2+^3C_3(3a^2)^3\\ =(3x^2-2ax)^3+3(9x^4-12ax^3+4a^2x^2)(3a^2)+3(3x^2-2ax)(9a^4)+27a^6\\ =(3x^2-2ax)^3+81a^2x^4-10a^2x^4-108a^3x^3+36a^4x^2+81a64x^2-54a^5x+27a^6\\ =(3x^2-2ax)^3+81a^2x64-108a^3x^3+117a^4x^2-54a^5x+27a^6.............(1)$ $\\$ Again by using Binomial Theorem, we obtain$\\$ $(3x^2-2ax)^3\\ =^3C_0(3x^2)^3-^3C_1(3x^2)^2(2ax)+^3C_2(3x^2)(2ax)^2-^3C_3(2ax)^3\\ =27x^6-3(9x^4)(2ax)+3(3x^2)(4a^2x^2)-8a^3x^3\\ =27x^6-54ax^5+36a^2x^4-8a^3x^3............(2)$ $\\$ From (1) and (2), we obtain $\\$ $(3x^2-2ax+3a^2)^3\\ =27x^6-54ax^5+36a^2x^4-8a^3x^3+81a^2x^4-108a^3x^3+117a^4x^2-54a^5x+27a^6\\ =27x^6-54ax^5+117a^2x^4-116a^3x^3+117a^4x^2-54a^5x+27a^6$