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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?

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In addition to having the wrong summation, you have a numerical error: $F_8=21$, not $24$. –  Brian M. Scott Jun 23 '13 at 1:18

2 Answers 2

up vote 5 down vote accepted

Hint: When we increment $n$ by $1$, we add two terms to the sum, not one.

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Grag! Stupid errors. Thank you! –  Emrakul Jun 23 '13 at 1:22
4  
You are welcome. The good thing is that you actually computed to see whether things made sense. Much more significant than the minor slips. –  André Nicolas Jun 23 '13 at 1:26

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 $$\sum_{j=2}^{2n}F_{j-1}F_j=F_{2n}^2$$

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