<s>[INST] If $f(n+1)=(-1)^{n+1}n-2f(n)$ and $f(1)=f(1986)$ find $f(1)+f(2)+f(3)+...+f(1985)$ I got a solution but it was really messy with f(1)=$\frac{1987}{2^{1985}+1}$ and an explicit formula for $f(x)$. Then the whole mess was a crocodile (With many $2^{1985}$ and $4^{1985}$ terms) when I solved it so I assume I made a mistake somewhere. (The answer must be simpler because it's an ARML problem)
 [/INST] Here's a fully worked version. The sum we are looking for is ($N = 1986$)
$$
\begin{align}
S = \sum_{n = 1}^{N - 1}f(n) &= \sum_{n = 1}^{N - 1}\left((- 1)^n(n-1) - 2f(n-1)\right)\\
&= \sum_{n = 1}^{N - 1}(-1)^n(n-1) - 2\sum_{n = 1}^{N - 1}f(n-1)\ .
\end{align}
$$
If we look at the first sum on the right side and write it out, knowing $N$ is even, we get
$$
\begin{align}
\sum_{n = 1}^{N - 1}(-1)^n(n-1) &= 0 + 1 - 2 + 3 - 4 + \dots + (N - 3) - (N - 2)\\
&=0 + (1 - 2) + (3 - 4) + \dots + ((N - 3) - (N - 2))\\
&=1 - \frac{N}{2}\ .
\end{align}
$$
Looking at the sum on the right side, if we substitute $n \rightarrow n + 1$ then
$$
\begin{align}
\sum_{n = 1}^{N - 1}f(n-1) = \sum_{n = 0}^{N - 2}f(n) = \sum_{n = 1}^{N - 1}f(n) + f(0) - f(N-1) = S + f(0) - f(N-1)\ .
\end{align}
$$
Again knowing $N$ is even in our case we can calculate
$$
\begin{gather}
f(1) = -2f(0)\quad \Rightarrow \quad f(0) = -\frac{1}{2}f(1)\ , \\
f(N) = -(N-1) - 2f(N-1)\quad \Rightarrow\quad f(N - 1) = -\frac{f(N) + (N - 1)}{2}\ .
\end{gather}
$$
Substituting into the equation above we have
$$
\sum_{n = 1}^{N - 1}f(n-1) = \sum_{n = 1}^{N - 1}f(n) + \frac{1}{2}\left(f(N) - f(1) - (N - 1)\right)\ .
$$
Substituting back into $S$, and noting that $f(1) = f(N)$, we finally get
$$
\begin{align}
S &= 1 - \frac{N}{2} - 2S + \left(f(1) - f(N)\right) + N - 1\\
\Rightarrow 3S &= \frac{N}{2}\\
\Rightarrow S &= \frac{N}{6}\ .
\end{align}
$$
</s>