Question

Find the coefficient of $x^2$ in $(x^2+\frac{1}{x^3}{)}^6$.

Answer 1

b'

Let $T_{r+1}$ be the term containing $x^2$.

$T_{r+1}={6_C}_r\left(x^2{)}^{6-r}\right(\frac{1}{x^3}{)}^r$

$={6_C}_r{x}^{12-5r}$

As the coefficient of x is 2.

$12-5r=2$

$r=2$

Therefore, coefficient of $x^2={6_C}_2=15$

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