Question

Find the coeficcient of the x4y3 term in the expansion of (2x+y)7

Answer 1

By the binomial theorem,

$\displaystyle (2x+y)^7 = \sum_{k=0}^7 \binom7k (2x)^{7-k} y^k=\sum_{k=0}^7 \binom7k 2^{7-k} x^{7-k} y^k$

where

$\dbinom nk=\dfrac{n!}{k!(n-k)!}$

is the binomial coefficient.

We get the x⁴y³ terms with k = 3, for which the coefficient would be

$\dbinom73 2^{7-3}=\dfrac{7!}{3! 4!} 2^4=\boxed{560}$