Question

$7^{2n}+2^{3n-3}{.3}^{n-1}$ is divisible by $25$ for any natural number $n\ge 1$. Prove that by mathematical

Answer 1

Step 1. Check if $P\left(i\right)$ is correct

We get $P\left(1\right)=7^2+2^0\times 3^0=49+1=50$

Which is divisible by $25$

So $P\left(1\right)$ is true.

Step 2. Let $P\left(K\right)$ is true, which means

$7^{2k}+3^{3k-3}{3}^{k-1}=25m$

When $m=integer$

Now check if $P(K+1)$ is True

Hence, $P\left(KH\right)=7^{2(K+H)}+2^{3(K+1)-3}\times 3^{(K+1)-1}$

$=(7^{2k}\times 49)+2^{(3k-3)}\times 8\times 3^{(k-1)}\times 3$

$=49\times 7^{2k}+24\left[2^{3k-3}{3}^{k-1)}\right]$

$=49\times 7^{2k}+24[25m-7^{2k}]$

$={49.7}^{2k}-{24.7}^{2k}+24.25m$

$=25[7^{2k}+24m]$

$=25\times$ some integers value

$\therefore$ $P(K+1)$ is true whenerver $P\left(K\right)$ is true.so using $PMI$, we can conclude $P\left(n\right)$ is true for all

$n\in N$