Write $\sum_{1}^{n} F_{2n-1} \cdot F_{2n}$ in a simpler form, where $F_n$ is the n-th element of the Fibonacci sequence? The exercise asks to express the following:
$\sum_{1}^{n} F_{2n-1} \cdot F_{2n}$
in a simpler form, not necessarily a closed one. The previous problem in the set was the same, with a different expression:
$\sum_{0}^{n} F_{n}^{2}$ which equals $F_{n} \cdot F_{n+1}$  
Side note:
I just started to work through an analysis book, my first big self-study effort. This problem appears in the introductory chapter with topics such as methods of proof, induction, sets, etc.

You want to watch those indices.  I think you mean $\sum_{k=0}^n F_k^2 = F_n F_{n+1}$ and $\sum_{k=1}^n F_{2k-1} F_{2k}$.  Hmm, it looks to me like this one can be expressed as a linear combination of $n$, $1$ and a certain Fibonacci number...
