Find the coefficient of x8 in the expansion of (x2-3)7
(n n-j) an-jxj
The Attempt at a Solution
Well, I know that
and apparently, j=8.
This is what confuses me. n-j is a negative number.. how would that even work?? My guess is that the x2 has to do with it somehow... I'm thinking because the x is squared, then n wouldn't be 7.
I wrote this on wolfram alpha because I was like.. "is this possible or is it just a typo?" and it gave me an answer!