 # Binomial Theorem

## Class 11 NCERT

### NCERT

1   Expand the expression $(1-2x)^5$

By using Binomial Theorem, the expression $(1-2x)^5$ can be expanded as $(1-2x)^5$$\\ {}^5C_{0}(1)^5 -{}^5C_1(1)^4(2x)+{}^5C_2(1)^3(2x)^2\\ -{}^5C_3(1)^2(2x)^3+{}^5C_4(1)^1(2x)^4-{}^5C_5(2x)^5\\ =1-5(2x)+10(4x)^2-10(8x^3)+5(16x^4)-(32x^5)\\ =1-10x+40x^2-80x^3+80x^4-32x^5 2 Expand the expression (\dfrac{2}{x}-\dfrac{x}{2})^5 ##### Solution : By using Binomial Theorem, the expression (\dfrac{2}{x}-\dfrac{x}{2})^5can be expanded as\\ (\dfrac{2}{x}-\dfrac{x}{2})^5\\ ={}^5C_0(\dfrac{2}{x})^5-{}^5C_1(\dfrac{2}{x})^4(\dfrac{x}{4})\\ +{}^5C_2(\dfrac{2}{x})^3(\dfrac{x}{2})^2\\ -{}^5C_3(\dfrac{2}{x})^2(\dfrac{x}{2})^3\\ +{}^5C_4(\dfrac{2}{x})(\dfrac{x}{2})^4-{}^5C_5(\dfrac{x}{2})^5$$\\$ $=\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 $(102)^5$

$102$ 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 $(99 ) ^5$

$99$ 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 $( x + 3 )^ 8$

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.

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$$\\ Thus, n=7 and r=3 25 Prove that the coefficient of x^n in the expansion of (1+x)^{2n} is twice the coefficient of x^nin the expansion of (1+x)^{2n-1} ##### Solution : It is known that (r+1)^{th} term, (T_{r+1}) in the binomial expansion of (a+b)^n is given by T_{r+1}=^nC_r a^{n-r} b^r . \\ Assuming that x^n occurs in the (r+1)^{th} term of the expansion of (1+x)^{2n},we obtain \\ T_{r+1}=^{2n}C_r (1)^{2n-r} (x)^r=^{2n}C_r(x)^r \\ Comparing the indices of x in x^n and in T_{r+1}, we obtain r=n \\ Therefore, the coefficient of x^n in the expansion of (1+x)^{2n} is \\ ^{2n}C_n=\dfrac{(2n)!}{n!(2n-n)!}=\dfrac{(2n)!}{n!n!}=\dfrac{(2n)!}{(n!)^2}.....(1) \\ Assuming that x^n occurs in the (k+1)^{th} term of the expansion of (1+x)^{2n-1},we obtain \\ T_{k+1}=^{2n}C_k(1)^{2n-1-k}(x)^k \\ Comparing the indices of x in x^n and in T_{k+1}, we obtain k=n \\ Therefore, the coefficient of x^ n in the expansion of (1+x)^{2n-1} is \\ ^{2n-1}C_n=\dfrac{(2n-1)!}{n!(2n-1-n)!}=\dfrac{(2n-1)!}{n!(n-1)!}\\ =\dfrac{2n.(2n-1)!}{2n.n!(n-1)!}=\dfrac{(2n)!}{2.n!n!}=\dfrac{1}{2}[\dfrac{(2n)!}{(n!)^2}]......(2) \\ From (1) and (2) , it is observed that \\ \dfrac{1}{2}(^{2n}C_n)=^{2n-1}C_n\\ \implies ^{2n}C_n=2(^{2n-1}C_n) \\ Therefore, the coefficient of x^n expansion of (1+x)^{2n} is twice the coefficient of x^n in the expansion of (1+x)^{2n-1}.$$\\$ 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 $\\$ $(3x^2-2ax+3a^2)^3\\ =27x^6-54ax^5+36a^2x^4-8a^3x^3+81a^2x^4-108a^3x^3+117a^4x^2-54a^5x+27a^6\\ =27x^6-54ax^5+117a^2x^4-116a^3x^3+117a^4x^2-54a^5x+27a^6$