How to prove? (Do not use mathematical induction) I would appreciate if somebody could help me with the following problem:
Q: Show that 
$$f_1=f_2=1, f_{n+2}=f_{n+1}+f_{n}(n\in \mathbb{N})~~~~ \Rightarrow ~~~~f_1f_2+f_2f_3+f_3f_4+\cdots+f_{2n-1}f_{2n}=f_{2n}^2$$
 A: The key idea of the classical proof is  telescopy. Push the recurrence back $\,1,\,$ so $\,f_0 = 0.\,$ Note that
$$\color{#c00}{f_{2k}^2 - f_{2k-2}^2} = (f_{2k}\!-f_{2k-2})(f_{2k}\!+f_{2k-2}) = f_{2k-1}(f_{2k}\!+f_{2k-2}) = \color{#c00}{f_{2k}f_{2k-1}\! + f_{2k-1}f_{2k-2}}$$
Summing the above from $\,k=1\,$ to $\,n,\,$ the left side telescopically cancels to $\,f_{2n}^2,\,$ yielding 
$$\begin{eqnarray} 
f_{2n}^2 &\,=\,& \qquad \color{#c00}{f_{2n}^2} &\color{#c00}-&\color{#c00}{f_{2n-2}^2} &+&\cdots &+&\quad f_4^2 &-& f_2^2 &+&\quad f_2^2 &-& f_0^2 \\
&\,=\,&   \color{#c00}{ f_{2n}f_{2n-1}} &\color{#c00}+&\color{#c00}{f_{2n-1}f_{2n-2}} &+&\cdots &+& f_4 f_3 &+& f_3 f_2 &+& f_2 f_1 &+& f_1 f_0\end{eqnarray}$$
Whether or not telescopy "uses induction" depends on how that term is defined.  One could inductively prove a lemma giving the general result for telescopic sums, then invoke that lemma, and that probably is a proof "not by induction".  Equivalently, one could deduce it by uniqueness of solutions of the associated linear difference equation (a recurrence  reformulation of telescopy). The lemma on telescopic sum evaluation (or the equivalent uniqueness theorem) has a very obvious one-line inductive proof (search on "telescopy" to find many prior answers that discuss such).
A: Hint: try rewriting the sum in a clever way to show what you want. To start, consider, $f_{2n}f_{2n-1} = f_{2n}(f_{2n}-f_{2n-2})$. Then see if you can't somehow make the series telescope somehow by making a similar (generalized) substitution. Then observe that $f_{1}f_{2} = f_{2}^{2}$... 
(I will note that the formal use of induction that you are used to seeing is makes this problem much easier, and in fact the hint I gave you does use induction - it just avoids the typical format you are used to seeing.)
A: EDIT: Sorry, I'm not sure if I didn't notice the full title or if it's been edited since. I would still strongly recommend you use proof by induction though unless it has been expressly forbidden.
I would strongly recommend using a proof by induction.
Show this is true when $n=1$ (should be simple)
Assume this is true for $n=k$. Now what terms do you need to add to the left hand side to make it look like the rule for $n=k+1$. Add these terms to the right hand side too. Can you now make the right hand side look like it should if the rule holds for $n=k+1$? Then you're done!
