Question

Prove that 21 divides $4^{n+1}+5^{2n-1}$whenever n is a positive integer.

Short answer

21 divides $4^{n+1}+5^{2n-1}$ whenever n is a positive integer

Step-by-step solution

Step: 1

If n=1,

$\begin{array}{r}{4}^{1+1}+5^{2-1}\\ =16+5=21\end{array}$

it is true for n=1.

Step: 2

Let P(k) be true.

$4^{k+1}+5^{2k-1}$

We need to prove that P(k+1) is true.

Step: 3

$\begin{array}{r}{4}^{(k+1)+1}+5^{(2k-1)+1}\\ =4\cdot 4^{k+1}+5^2\cdot 5^{2k-1}\\ =4\cdot 4^{k+1}+25\cdot 5^{2k-1}\\ =4\cdot 4^{k+1}+(21+4)\cdot 5^{2k-1}\\ =4\cdot 4^{k+1}+21\cdot 5^{2k-1}+4\cdot 5^{2k-1}\\ =21\cdot 5^{2k-1}+4\cdot \left(4^{k+1}+5^{2k-1}\right)\end{array}$

It is divisible by 21.

It is true for P(k+1) is true.