For (n ∈ N, x^ n+1 +(x+1)^ 2 n-1 ) is divisible by

For $n=1$, we have; (x^{n+1}+(x+1)^{2 n-1}=x²+(x+1)=x²+x+1 \text{, }) which is divisible by $x²+x+1$ For $n=2$, we have; $x^{n+1}+(x+1)^{2 n-1}$ (=x³+(x+1)^3=(2 x+1)\left(x^2+x+1\right)\text{, }) which is divisible by $x²+x+1$.