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$\\$ $(x+\dfrac{1}{x})^6\\ ={}^6C_0(x)^6+{}^6C_1(x)^5(\dfrac{1}{x})+\\ {}^6C_2(x)^4(\dfrac{1}{x})^2+{}^6C_3(x)^3(\dfrac{1}{x})^3\\ +{}^6C_4(x)^2(\dfrac{1}{x})^4+{}^6C_5(x)(\dfrac{1}{x})^5+{}^6C_6(\dfrac{1}{x})^6\\ =x^6+6(x)^5(\dfrac{1}{x})+15(x)^4(\dfrac{1}{x^2})+\\ 20(x)^3(\dfrac{1}{x^3})+15(x)^2(\dfrac{1}{x^4})6(x)(\dfrac{1}{x^5})+\dfrac{1}{x^6}\\ =x^6+6x^4+15x^2+20+\dfrac{15}{x^2}+\dfrac{6}{x^4}+\dfrac{1}{x^6}$