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.
 A: In this paper, you may find equation 18 useful. 
A: 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...
A: You can try to use closed form (Cauchy-Binet formula) for Fibonacci and Lucas sequence.
$F_n=\frac{\alpha^n-\beta^n}{\sqrt{5}}$
$L_n=\alpha^n+\beta^n$
where $\alpha$ and $\beta$ are the roots of $x^2-x-1$. It is useful to notice that $\alpha+\beta=1$ and $\alpha.\beta=-1$.
Thus $F_{2k}F_{2k-1}=\frac{(\alpha^{2k}-\beta^{2k})(\alpha^{2k-1}-\beta^{2k-1})}5 = \frac{\alpha^{4k-1}-\beta^{4k-1}}5 - \frac{(\alpha\beta)^{2k-1}(\alpha+\beta)}5 = \frac{L_{4k-1}+1}5$. (I believe this is a special case of equations (18) from picakhu's post.)
Thus it remains to calculate $\sum L_{4k-1}=\sum (\alpha^k+\beta^k)$, which is basically using geometric progressions and algebraic manipulations. (The identities $\alpha+\beta=1$ and $\alpha.\beta=-1$ might be handy.)
