Question

find the coefficient x^15 in the expansion of (x^3+2/x^2)^10

Answer 1

Answer:

        960

Step-by-step explanation:

By the binomial theorem, the terms in the expansion take the form

$\displaystyle\binom{10}{k}(x^3)^{10-k}\bigl(\tfrac2{x^2}\bigr)^k = 2^k\binom{10}{k}x^{30-5k}$

The term corresponding to x¹⁵ is the one with k=3.

The coefficient is then

$\displaystyle2^3\binom{10}{3}=\frac{2^3\times10\times9\times8}{3\times2\times1}=960$