I have been assigned a very thought-provoking functional computation problem (to be completed $without$ a calculator) which has left me essentially stumped—that is, I really can't come up with an efficient method of solving this problem without simply plugging in attractive values.
The question is as follows: If $f(n+1) = n(-1)^{n+1}- 2f(n)$ for integral $n>1$, and $f(1)=f(1986)$, compute $$f(1) + f(2) + f(3) + f(4) + ... + f(1985).$$ Many thanks for any words of advice or hints.
Hint: Let $S = \displaystyle\sum_{k = 1}^{1985}f(k)$. Then, since $f(1) = f(1986)$, we have:
$S = \displaystyle\sum_{k = 1}^{1985}f(k)$ $= \displaystyle\sum_{k = 2}^{1986}f(k)$ $= \displaystyle\sum_{n = 1}^{1985}f(n+1)$ $= \displaystyle\sum_{n = 1}^{1985}\left(n(-1)^{n+1} - 2f(n)\right)$
$= \left(\displaystyle\sum_{n = 1}^{1985}n(-1)^{n+1}\right) - 2\left(\displaystyle\sum_{n = 1}^{1985}f(n)\right)$ $= \left(\displaystyle\sum_{n = 1}^{1985}n(-1)^{n+1}\right) - 2S$.
Therefore, $3S = \displaystyle\sum_{n = 1}^{1985}n(-1)^{n+1}$.