Recurrence equation for $(-1)^k k$

Question

In a project of mine I came across the recurrence relation $$ a_{n+1} = 1 -(n+1)\sum_{k=1}^n{\frac{a_k}{n-k+1}\binom{n}{k}},\quad a_1=2; $$

From calculating the first few terms it seems obvious that $$ a_n = (-1)^{n+1}(n+1),\quad n\geqslant2, $$ and, once guessed, this solution can be verified by induction.

My question is if there is an easy way to arrive at this solution without guessing. Generating functions work but seem unnecessarily unwieldy for such an easy-looking recurrence.

Answer 1

With $a_0=-1$, your recurrence becomes

$$\sum_{k=0}^na_k\binom nk =f(n)=\begin{cases}-1 &\text{ if $n=0$,}\\ 1 &\text{ if $n=1$,}\\ 0 &\text{ if not.}\end{cases}$$

This implies the inverse relation $$a_n=\sum_{k=0}^n(-1)^{n-k}\binom{n}kf(k)=(-1)^{n}(-1)+(-1)^{n-1}n =(-1)^{n+1}(n+1)$$ as desired.

Answer 2

$$ a_{n+1} = 1-\sum_{k=1}^n{\binom{n+1}{k}a_k} $$ maybe use formula $$\sum_{k=1}^n{(-1)^k k\binom{n}{k}}=0$$