Writing a *finite* power series as a rational expression

Question

$$ \sum_{k=0}^{14}(-3)^kx^{2k} $$ I know that to get

$$ \sum_{k=0}^\infty(-3)^kx^{2k} = \frac3{3+x^2} $$

Answer 1

Since $\frac1{1-x} =\sum_{n=0}^{\infty} x^n $, we have $x^m\frac1{1-x} =\sum_{n=0}^{\infty} x^mx^n =\sum_{n=0}^{\infty} x^{m+n} =\sum_{n=m}^{\infty} x^{n} $.

Subtracting, $\sum_{n=0}^{m-1} x^{n} =\sum_{n=0}^{\infty} x^{n}-\sum_{n=m}^{\infty} x^{n} =\frac1{1-x}-\frac{x^m}{1-x} =\frac{1-x^m}{1-x} $.

A nice thing about this result is that it is true for any $x \ne 1$, even though the original sum is valid only for $-1 \le x < 1$.

Therefore $\sum_{k=0}^{14}(-3)^kx^{2k} =\sum_{k=0}^{14}(-3x^2)^{k} =\frac{1-(-3x^2)^{15}}{1-(-3x^2)} $ by replacing $m$ with $15$ and $x$ with $-3x^2$.

Answer 2

Hint: can you do $$ \sum_{k=15}^{\infty} (-3)^kx^{2k}? $$

\begin{align} =((-3)x^2)^{15} \sum_{m=0}^{\infty} (-3)^m x^{2m} \end{align}