By Binomial Theorem
(a+b)n=(0n)anb0+(1n)an−1b1+...+
+(kn)akbn−k+...+(nn)a0bnWe have a=−x,b=2y,n=16.
We find the coefficient of x3y13
k=3,(kn)akbn−k=(316)(−x)3(2y)16−3=
=−3!(16−3)!16!(8192)x3y13=−560(8192)x3y13=
=−4587520x3y13
The coefficient of x3y3 in the expansion of (−1x+2y)16 is −4587520.
Comments
Leave a comment