TITLE: Error in Fibonacci recurrence proof by induction?
QUESTION [7 upvotes]: I'm working on a problem from a number theory book (Number Theory by George E. Andrews - problem 1-1-11). The text reads:

Prove: $\displaystyle F_1F_2+F_2F_3+F_3F_4+\ldots+F_{2n-1}F_{2n}=F_{2n}^2$

I started by setting up a summation:
$$\sum_{j=1}^{n}F_{2j-1}F_{2j}=F_{2n}^2$$
From this, I worked inductively:
$$\begin{align*}
\sum_{j=1}^{n+1}F_{2j-1}F_{2j}&=F_{2n+2}^2\\
F_{2(n+1)-1}F_{2(n+1)}+\sum_{j=1}^{n}F_{2j-1}F_{2j}&=\ldots\\
F_{2n+1}F_{2n+2}+F_{2n}^2&=F_{2n+2}^2
\end{align*}$$
This doesn't mathematically work, though. Let $n=3$, and evaluate. 
$$\begin{align*}F_7F_8+F_6^2&=F_8^2\\13\cdot21+8^2&=21^2\\337&\neq441\end{align*}$$
Where did I go wrong? Any ideas?

REPLY [4 votes]: Hint: When we increment $n$ by $1$, we add two terms to the sum, not one.

REPLY [2 votes]: Your summation is incorrect.  It includes only the terms odd*even, but not the terms even*odd like $F_2F_3$  (also the indices should be $j$, not $n$).  The correct way to write it is $$\sum_{j=2}^{2n}F_{j-1}F_j=F_{2n}^2$$