Question

Prove that ${\mathbf{f}}_{\mathbf{1}}^{\mathbf{2}}\mathbf{+}{\mathbf{f}}_{\mathbf{2}}^{\mathbf{2}}\mathbf{+}\mathbf{\dots }\mathbf{\dots }\mathbf{.}\mathbf{.}{\mathbf{f}}_{\mathbf{n}}^{\mathbf{2}}\mathbf{=}{\mathbf{f}}_{\mathbf{n}}{\mathbf{f}}_{\mathbf{n}\mathbf{+}\mathbf{1}}$where n is a positive integer

Short answer

${\mathrm{f}}_1^2+{\mathrm{f}}_2^2+\dots \dots ..{\mathrm{f}}_{\mathrm{n}}^2={\mathrm{f}}_{\mathrm{n}}{\mathrm{f}}_{\mathrm{n}+1}$ whenever n is a positive integer.

Step-by-step solution

The definition of Fibonacci numbers

By the definition of Fibonacci numbers

$\begin{array}{r}\mathrm{f}\left(0\right)=0\\ \mathrm{f}\left(1\right)=1\\ {\mathrm{f}}_{\mathrm{n}}={\mathrm{f}}_{\mathrm{n}-1}+{\mathrm{f}}_{\mathrm{n}-2}\text{when}\mathrm{n}\ge 2\end{array}$

To prove that f12+f22+……fn2=fnfn+1 whenever n is a positive integer

The proof is given by INDUCTION:

Let P(n) be ${\mathrm{f}}_1^2+{\mathrm{f}}_2^2+\dots \dots {\mathrm{f}}_{\mathrm{n}}^2={\mathrm{f}}_{\mathrm{n}}{\mathrm{f}}_{\mathrm{n}+1}$

Inductive step: Assume that P(k) is true.

To prove that P(K+1) is also true.

Therefore, it can be written as:

$\begin{array}{r}{\mathrm{f}}_1^2+{\mathrm{f}}_2^2+\dots {\mathrm{f}}_{\mathrm{k}}^2+{\mathrm{f}}_{\mathrm{k}+1}^2={\mathrm{f}}_{\mathrm{k}}{\mathrm{f}}_{\mathrm{k}+1}+{\mathrm{f}}_{\mathrm{k}+1}^2\\ ={\mathrm{f}}_{\mathrm{K}+1}\left({\mathrm{f}}_{\mathrm{k}}+{\mathrm{f}}_{\mathrm{k}+1}\right)\\ ={\mathrm{f}}_{\mathrm{K}+2}{\mathrm{f}}_{\mathrm{k}+1}\\ ={\mathrm{f}}_{\mathrm{k}+1}{\mathrm{f}}_{(\mathrm{k}+1)+1}\end{array}$

Hence, it is proved that P (k + 1) is true.

Thus, it is proved by the principle of mathematical induction that:$f_1^2+f_2^2+\dots f_n^2=f_n{f}_{n+1}$whenever n is a positive integer.