Question

Prove that ${\mathbf{f}}_{\mathbf{0}}{\mathbf{f}}_{\mathbf{1}}\mathbf{+}{\mathbf{f}}_{\mathbf{1}}{\mathbf{f}}_{\mathbf{2}}\mathbf{+}\mathbf{\dots }\mathbf{\dots }\mathbf{.}{\mathbf{f}}_{\mathbf{2}\mathbf{n}\mathbf{-}\mathbf{1}}{\mathbf{f}}_{\mathbf{2}\mathbf{n}}\mathbf{=}{\mathbf{f}}_{\mathbf{2}\mathbf{n}}^{\mathbf{2}}$where n is a positive integer.

Short answer

${\mathrm{f}}_0{\mathrm{f}}_1+{\mathrm{f}}_1{\mathrm{f}}_2+\dots \dots .{\mathrm{f}}_{2\mathrm{n}-1}{\mathrm{f}}_{2\mathrm{n}}={\mathrm{f}}_{2\mathrm{n}}^2$ whenever n is a positive integer.

Step-by-step solution

The definition of Fibonacci numbers

By the definition of Fibonacci numbers

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

To prove that f0f1+f1f2+……f2n−1f2n=f2n2 whenever n is a positive integer

The proof is given by INDUCTION:

Let P(n) be $f_0{f}_1+f_1{f}_2+\dots \dots f_{2n-1}{f}_{2n}=f_{2n}^2$

Inductive step:

Assume that P(k) is true.

Therefore, we have $f_0{f}_1+f_1{f}_2+\dots \dots f_{2k-1}{f}_{2k}=f_{2k}^2$

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

Therefore, it can be written as:

$\begin{array}{r}{f}_0{f}_1+f_1{f}_2+\dots .f_{2(k+1)-1}{f}_{2(k+1)}=f_0{f}_1+f_1{f}_2+\dots f_{2k+1}{f}_{2k+2}\\ =f_{2(k+1)}^2\end{array}$

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

Thus, it is proved by the principle of mathematical induction that:

$f_0{f}_1+f_1{f}_2+\dots .f_{2n-1}{f}_{2n}={f_{2n}}^2$whenever n is a positive integer.