2016/103) if p is natural number show that $p^{n+1} + (p+1)^{2n-1}$ is divisible by $p^2+p+1$

let $f(n) = p^{n+1} + (p+1)^{2n-1}$
we shall prove it principle of mathemetical induction
for n = 1 we have  $f(1) = p^2 + p+ 1 $ and obviously it is divsible by $p^2+p+1$
so we have shown it for base step
let it be dvisible for n = k so $p(k)$ is divisible by $p^2+p+1$
or $p^{k+1} + (p+1)^{2k-1} =  q(p^2+p+1)$
Now we need to show that it is true for k+ 1
$p^{k+2} + (p+1)^{2k+1} = p^(k+2) + (p+1)^2 (p+1)^{2k-1}$
$ = p^{k+2}+ (p^2 + 2p + 1) (p+1)^{2k-1}$
$= p^{k+2} + p((p+1)^{2k-1}) + (p^2 + 2p + 1)((p+1)^{2k-1})$
$=  p(p^{k+1} + (p+1)^{2k-1} + (p^2 + 2p + 1)((p+1)^{2k-1})$
$ = p(q(p^2+p+1)) + p^2 + 2p + 1)((p+1)^{2k-1})$
which is divisible by $p^2+p+1$ hence induction step is proved.
hence proved