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  4. For n ∈ N, xn+1+(x+1)2n-1 is divisible by

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Q. For $n \in N, x^{n+1}+\left(x+1\right)^{2n-1}$ is divisible by

Principle of Mathematical Induction

Solution:

For $n = 1,$ we have
$x^{n+1} + (x + 1)^{2n-1} = x^2 + (x + 1) = x^2 + x + 1,$
which is divisible by $x^2 + x + 1$
For $n = 2$, we have
$x^{n+1} + (x + 1)^{2n-1} = x^3 + (x + 1)^3 = (2x + 1) (x^2 + x + 1),$ which is divisible by $x^2 + x + 1.$