## Related questions with answers

Prove the following statements with either induction, strong induction or proof by smallest counterexample. Concerning the Fibonacci sequence, prove that $\sum_{k=1}^{n} F_{k}^{2}=F_{n} F_{n+1}$.

Solution

Verified$\text{\underline{Remember:}}$ The $\text{\textcolor{#c34632}{Fibonacci numbers}}$ are defined to be $F_1 = 1 , F_2 = 1$ and $F_n = F_{n-1} + F_{n-2}$ for $n > 2$.

$\textbf{Proof by Mathematical induction}$

For $n=1$: L.H.S $=\sum_{k=1}^{1}{F_k}^2={F_1}^2=1$, R.H.S $=F_1F_2=1 \cdot 1=1$ Thus it is true for $n=1$.

Assume it is true for $n=t$ then $\color{#4257b2}\sum_{k=1}^{t}{F_k}^2=F_tF_{t+1}\qquad \cdots (1)$

Now for $n=t+1$:

$\begin{align*} &\sum_{k=1}^{t}{F_k}^2\\ &=\sum_{k=1}^{t+1}{F_k}^2+{F^2}_{t+1}\\ &\overset{{\color{#4257b2}(1)}}{=}F_tF_{t+1}+{F^2}_{t+1}\\ &=F_{t+1}(F_t+F_{t+1})\\ &=F_{t+1}F_{t+2} \end{align*}$

Thus it is true for $n=t+1$.

Therefore by mathematical induction $\sum_{k=1}^{n}{F_k}^2=F_nF_{n+1}$ for every integer $n \in \Bbb N \qquad \blacksquare$

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