Solution
= Solution
Substitute $r=-x^2$ into the <geometric series>, without taking any <derivatives>:
$$
\boxed{\frac1{1+x^2}=\sum_{k=0}^{\infty}(-1)^k x^{2k}
=1-x^2+x^4-\cdots,\qquad |x|<1.}
$$
Thus the coefficient of $x^n$ in the <Taylor series> is $(-1)^{n/2}$ when $n$ is even and zero when $n$ is odd. If nonzero terms are numbered from one, the $n$th term is $(-1)^{n-1}x^{2n-2}$.