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I want to prove the below:

$$\sum_{i=1}^n f_i^2 = f_n \times f_{n+1}$$

The example of $n = 1$ is trivial:

$$\begin{align} \sum_{i=1}^1 f_i^2 &= f_1^2 \\ &= 1^2 \\ &= 1 \times 1 \\ &= f_1 \times f_2 \end{align}$$

And working through $n=2$:

$$\begin{align} \sum_{i=1}^2 f_i^2 &= f_1^2 + f_2^2 \\ &\stackrel{?}{=} f_2 \times f_3 \end{align}$$

Using the definition of a Fibonacci number, I can work backwards from $f_2 \times f_3$:

$$\begin{align} f_2 \times f_3 &= (f_{2-1}+f_{2-2}) \times (f_{3-1}+f_{3-2}) \\ &= (f_1 + f_0) \times (f_2 + f_1) \\ &= f_1f_2 + f_1f_1 + f_0f_2 + f_0f_1 \\ &= f_1f_2 + f_1^2 \end{align}$$

Since $f_2 = 1$, $f_1f_2 = f_2f_2 = f_2^2$, thus $f_2 \times f_3 = f_1^2 + f_2^2$.

But how could I generalize this for any $n$ and not show examples for $n=3$, $n=4$, etc?

gator
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1 Answers1

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Use Mathematical Induction. If $n=1$, the identity is true. Suppose $n=k$ it is true; namely $$ \sum_{i=1}^k f_i^2 = f_k \times f_{k+1}. $$ Now for $n=k+1$, \begin{eqnarray} \sum_{i=1}^{k+1} f_i^2 &=& \sum_{i=1}^k f_i^2+f_{k+1}^2\\ &=&f_k \times f_{k+1}+f_{k+1}^2\\ &=&f_{k+1}(f_k+f_{k+1})\\ &=&f_{k+1}f_{k+2}\\ &=&f_{k+1}f_{(k+1)+1} \end{eqnarray} which implies that the identity is true for $n=k+1$. So $$\sum_{i=1}^n f_i^2 = f_n \times f_{n+1}$$ for any $n$.

xpaul
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