Question

Coefficient of x in the expansion of $(1-2x^3+3x^5){(1+\frac{1}{x})}^8$ is

• A. 154
• B. 164
• C. 146
• D. 156

Answer 1

The correct option is A 154

$=(1–2x^3+3x^5){(1+\frac{1}{x})}^8$

$=(1–2x^3+3x^5)(1+\frac{8}{x}+\frac{28}{x^2}+\frac{56}{x^3}+\frac{70}{x^4}+\frac{56}{x^5}+\frac{28}{x^6}+\frac{8}{x^7}+\frac{1}{x^8})$

For coefficient of $x$,

$=-2x^3\times \frac{28}{x^2}+3x^5\times \frac{70}{x^4}$

$=-56x+210x=154x$

Thus the coefficient of $x$ is $154$.