Question

Prove that ${\mathbf{f}}_{\mathbf{1}}\mathbf{+}{\mathbf{f}}_{\mathbf{3}}\mathbf{+}\mathbf{\dots }\mathbf{\dots }{\mathbf{f}}_{\mathbf{2}\mathbf{n}\mathbf{-}\mathbf{1}}\mathbf{=}{\mathbf{f}}_{\mathbf{2}\mathbf{n}}$ where n is a positive integer.

Short answer

${\mathrm{f}}_1+{\mathrm{f}}_3+\dots \dots {\mathrm{f}}_{2\mathrm{n}-1}={\mathrm{f}}_{2\mathrm{n}}$whenever n is a positive integer

Step-by-step solution

The definition of Fibonacci numbers

By the definition of Fibonacci numbers

$\begin{array}{r}\mathrm{f}\left(0\right)=0\\ \mathrm{f}\left(3\right)=1\\ {\mathrm{f}}_{\mathrm{n}}={\mathrm{f}}_{2\mathrm{n}-1}\text{when}\mathrm{n}\ge 1\end{array}$

To prove that f1+f3+……f2n−1=f2n  whenever n is a positive integer

The proof is given by INDUCTION:

Let P(n) be $f_1+f_3+\dots f_{2n-1}=f_{2n}$

Here, the sequence of Fibonacci numbers is defined by the relation:

$F\left(n\right)=F(n-1)+F(n-2)$

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

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

Therefore, it can be written as:

$\begin{array}{r}{f}_1+f_3+\dots .f_{2k-1}=f_{2(k+1)}\\ =f_{2k+2}\\ =f_{2k+1}{f}_{2n}\\ =f_{2k+1}+f_{2k-1}+\dots \dots f_3+f_1\\ =f_{2k}\end{array}$

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

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

$f_1+f_3+\dots \dots f_{2n-1}=f_{2n}$whenever n is a positive integer.