Let $f:\mathbb{N} \rightarrow \mathbb{N}$ such that $f(1)=2011$ and $f(n)= \frac{1}{n^2 -1}(f(1)+f(2)+...+f(n-1))$ for $n \ge 2$. Calculate $f(2011)$

Question

Being $f:\mathbb{N} \rightarrow \mathbb{N}$ such that $f(1)=2011$ and $f(n)= \frac{1}{n^2 -1}(f(1)+f(2)+...+f(n-1))$ for $n \ge 2$. Calculate $f(2011)$

When calculating $f(n)$ I need to consider the sum of the previous elements, from $2011$ to $f(n-1)$. But when considering each of these, I need again the the sum of the previous $f(n)$. I've never met a problem like this. It seems like a sequence by recursion: are there any rules when approaching problems like this?

Answer 1

Hint: let $g(n) = \sum_{k=1}^n f(k)$. Then we have $f(2011) = g(2011) - g(2010)$ and

$$g(n) = g(n-1) + f(n)$$

$$g(n) = g(n-1) + \frac{1}{n^2 - 1}g(n-1)$$

$$g(n) = \frac{n^2}{n^2 - 1}g(n-1)$$

Can you take it from here?