Expand $$(x+ \dfrac{1}{x})^6$$

Asked by Pragya Singh | 2 years ago |  85

##### Solution :-

$$^6C_0(x)^6+ ^6C_1(x)^5(\dfrac{1}{x})+ ^6C_2(x)^4(\dfrac{1}{x})^2$$

$$+^6C_3(x)^3+(\dfrac{1}{x})^3+^6C_4(x)^2(\dfrac{1}{x})^4+$$

$$^6C_5(x)(\dfrac{1}{x})^5+^6C_6(\dfrac{1}{x})^6$$

$$x^6+6(x)^5(\dfrac{1}{x})+15(x)(\dfrac{1}{x^2})$$

$$+20(x)^3(\dfrac{1}{x^3}) +15(x)(\dfrac{1}{x^2})+20(x)^3(\dfrac{1}{x^3})$$

$$+15(x)^2(\dfrac{1}{x^4})+6(x)(\dfrac{1}{x^5})+\dfrac{1}{x^6}$$

$$x^6+6x^4+15x+20+\dfrac{15}{x^2}+\dfrac{6}{x^4}+\dfrac{1}{x^6}$$

Answered by Abhisek | 2 years ago

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