Question

If $P\left(n\right):2^{3n}+a$ is divisible by $7(n\in N),$ then the minimum positive value of $a$ for which $P\left(n\right)$ is true, is

Answer 1

$P\left(n\right)=2^{3n}+a$ is divisible by $7$
As, $P\left(n\right)$ is true for all natural numbers of $n.$
it is also true for $n=1.$
For $n=1,P\left(1\right)=8+a$ is divisible by $7$ , if $a=-8,-1,6$
So, minimum positive value of $a=6$
now check for $P\left(n\right):2^{3n}+6$ is also true for $n=n+1$
$P(n+1):2^{3n+3}+6=8(7+1{)}^n+6$ is also divisible by $7,$ is true
So, $P\left(n\right)$ is $2^{3n}+6.$