Find the coefficient of x in the expansion of $$(1 – 2x^3 + 3x^5) (1 + \dfrac{1}{x})^8$$

Asked by Sakshi | 1 year ago |  85

##### Solution :-

If x occurs at the (r + 1)th term in the given expression.

Then, we have:

(1 – 2x3 + 3x5) (1 + $$\dfrac{1}{x}$$)8

So, ‘x’ occurs in the above expression at -2x3.8C2 ($$\dfrac{1}{x^2}$$) + 3x5.8C4 ($$\dfrac{1}{x^4}$$)

Coefficient of x =$$-2 (\dfrac{8!}{(2!6!) })+ 3 (\dfrac{8!}{(4! 4!)})$$

= -56 + 210

= 154

Answered by Sakshi | 1 year ago

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