which series is the function $\frac{1}{1-6x^2}$ generating? I think it should be $f(n)=6^n$
2026-03-29 12:04:51.1774785891
which series is the function $\frac{1}{1-6x^2}$ generating?
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1
$$\frac{1}{1-\alpha}=\sum_{n=0}^\infty \alpha^n$$
is the foundation. Apply with $\alpha=6x^2$ and we get
$$\frac{1}{1-6x^2}=\sum_{n=0}^\infty (6x^2)^n = \sum_{n=0}^\infty 6^n x^{2n}$$
so this corresponds to the sequence $1, 0, 6, 0, 36, \ldots$, or in a formula
$$a_n = \begin{cases} 0 & n \text{ odd }\\ 6^{\frac{n}{2}} & n \text{ even } \end{cases}$$
if you prefer.