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=(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)=(7!)/(3! 4!) 2^4=\boxed{560}`