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