# 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 $(99 ) ^5$