I'd like some clarification on how index substitution works.
Evaluating Combination Sum $\sum{n+k\choose 2k} 2^{n-k}$ <- in this example, the answer by @Leucippus involves switching n -> n+k in the summation in the 2nd step. I'd like to know how to apply this index substitution under different contexts.
The basic rule is that $$\sum_{n=a}^b f_n = \sum_{n=a-k}^{b-k} f_{n+k}.$$
For the linked answer, here are some additional steps for clarity, where $f_n=\binom{n+k}{2k}2^{-k}t^n$: \begin{align} \sum_{n=0}^\infty S_{n} \frac{t^n}{2^n} &= \sum_{n=0}^\infty \sum_{k=0}^n \binom{n+k}{2k} 2^{n-k} \frac{t^n}{2^n} \\ &= \sum_{n=0}^\infty \sum_{k=0}^n \binom{n+k}{2k} 2^{-k} t^n \\ &= \sum_{0 \le k \le n \le \infty} \binom{n+k}{2k} 2^{-k} t^n \\ &= \sum_{k=0}^\infty \sum_{n=k}^\infty \binom{n+k}{2k} 2^{-k} t^n \\ &= \sum_{k=0}^\infty \sum_{n=k-k}^\infty \binom{(n+k)+k}{2k} 2^{-k} t^{n+k} \\ &= \sum_{k=0}^\infty \sum_{n=0}^\infty \binom{n+2k}{2k} 2^{-k} t^{n+k} \\ &= \sum_{n=0}^\infty \sum_{k=0}^\infty \binom{n+2k}{2k} 2^{-k} t^{n+k} \end{align}
