Question

What is the coefficient of x^3 y^13 in the expansion of (−1x+2y)^16?

Answer 1

According to Newton's Binomial Theorem:

$(a+b)^n=\sum_{k=0}^n \frac{n!}{k!(n-k)!}a^kb^{n-k}$

In our case $a=-1x, b= 2y, n=16, k=3,n-k=13.$

Therefore, the coefficient of $x^3 y^{13}$ is

$\frac{16!}{3!13!}(-1)^32^{13}=-\frac{16\cdot 15\cdot 14\cdot 13!}{3!13!}2^{13}=-\frac{16\cdot 15\cdot 14\cdot }{6}2^{13}=$

$=-8\cdot 5\cdot 14\cdot 2^{13}=-5\cdot 7\cdot 2^{17}=-4,587,520$