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)
If $f(n+1)=(-1)^{n+1}n-2f(n)$ and $f(1)=f(1986)$ find $f(1)+f(2)+f(3)+...+f(1985)$
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contest-math
functional-equations
4 Answers
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Replace $1986$ by $N$. Finding a more general formula will be easier than doing messy arithmetic.
The general solution of this inhomogeneous linear equation is $$ f(n) = (n+1) (-1)^{n+1} + c (-2)^n $$ for arbitrary $c$. In order for $f(1) = f(N)$, where $N$ is even, you need $$c = \frac{N+3}{2^{N}+2} $$
Then $$\sum_{n=1}^{N-1} f(n) = - \frac{2^{N}+2}{3} c + \frac{N}{2} + 1 = \frac{N}{6}$$
In this case, your answer is $1986/6 = 331$.
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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
$$ \begin{align} S &= 1 - \frac{N}{2} - 2S + \left(f(1) - f(N)\right) + N - 1\\ \Rightarrow 3S &= \frac{N}{2}\\ \Rightarrow S &= \frac{N}{6}\ . \end{align} $$
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Idea : Write it for each n and on adding uptil 1985, we get : $ 3(\sum _{i=2}^{1985} f(i)) + 2f(1) + f(1986) = \sum_{j=1}^{1985} (-1)^{j+1} j $ .So you get the sum of all $f(i)$'s till 1985. I don't know how to get the last term as a value.
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Lets tackle this in a sighly more general case
Suppose $f(2k) = f(1)$
base case $k=1$
$\sum_\limits{i=1}^{2k-1} f(i)$
$f(i) = (-1)^i (2^i-i-1 - 2^{i-1} f(1))\\ f(2k) = 2^{2k}-2k-1 - 2^{2k-1} f(1)= f(1)\\ f(1) = \frac {2^{2k} - 2k - 1}{2^{2k-1} +1}$
$\sum_\limits{i=1}^{2k-1} (-2)^i = -\frac{2^{2k}+2}{3}\\ -\sum_\limits{i=1}^{2k-1} (-1)^i(i) = k\\ -\sum_\limits{i=1}^{2k-1} (-1)^i(1) = 1\\ -\sum_\limits{i=1}^{2k-1} (-1)^i 2^{2k-1}f(1) = \frac{(2^{2k-1}+1)}{3}f(1)=\frac{2^{2k}-2k-1}{3}\\ $
$\sum_\limits{i=1}^{2k-1} f(i) = \frac {k}{3}$
$\sum_\limits{i=1}^{1985} f(i) = 331$