Question

$5^{2n}-1$ is divisible by 24 for all n $ϵ$ N.

Answer 1

Let P(n) : $5^{2n}-1$ is divisible by 24

For n = 1

$5^2-1=24$

Which is divisible by 24

$\Rightarrow$ P(n) is true for n = 1

Let P(n) is true for n = k

$\Rightarrow (5^{2k}-1)$ is divisible by 24

$\Rightarrow 5^{2k}-1=24\lambda$ .......(1)

We have to show that,

$5^{2k}-1$ is divisible by 24

$5^{2(k+1)}-1=24\mu$

Now,

$5^{2(k+1)}-1$

$=5^{2k}{.5}^2-1$

$={25.5}^{2k}-1$

$=25(24\lambda +1)-1$

[Using equation (1)]

$=25.24\lambda +24$

$=24(25\lambda +1)$

$=24\mu$

$\Rightarrow$ P(n) is true for n = k + 1

$\Rightarrow$ P(n) is true for all n $ϵ$ N by PMI.