Find the coefficient of a negative indexed $x$ in a series expansion

Question

Find the coefficient of x-10 in the expansion:
(2-1/x2)8

ANS: -448

I've tried using the General Term formula and got stuck at -x^2r = x^-10. Also, I tried expanding the equation but it doesn't look like its leading me somewhere.

Answer 1

You've got the right idea. We know $$(a+b)^8=\sum_{k=0}^{8}{8\choose k}a^{8-k}b^k$$ so we just need to see what happens when $a=2, b=-x^{-2}.$ If order to get the $x^{-10}$ term, we need $k=5$ since $(x^{-2})^5=x^{10}.$ Then the coefficient will be $${8\choose5}2^{8-5}(-1)^5=-8\cdot56=-448$$

Answer 2

You have eight factors of $(2-\frac 1{x^2})$ You need to choose $5$ of them to supply $\frac 1{x^2}$ to get the right power of $x$, so $3$ of them give a $2$. There are ${8 \choose 5}=56$ ways to choose the five and the three factors of $2$ give a factor $8$. The answer is $$-56 \cdot 8=-448$$