If p is a fixed positive integer, prove by induction that `p^(n +1) + (p + 1)^(2n - 1)` is divisible by `P^2+ p +1` for all `n in N`.

Options

A. P
B. `P^2+P`
C. `P^2+P+1`
D. `P^2-1`

Explanation

For `n=1`, we get
`P^n+1)+(P+1)^(2n-1)=P^(2)+(P+1)^(1)=P^(2)+P+1`,
which is divisible by `P^2+P+1`, so result is true for `n=1`.
Let us assume that the given result is true for `n=m in N`.
i.e., `P^(m+1)+(P+1)^(2m-1)` is divisible by `P^2+P+1`.
i.e., `P^(m+1)+(P+1)^(2m-1)=k(P^2+P+1),forall k in N` ....(i)
Now , `P^((m+1)+1)+(P+1)^(2(m+1)-1)`
`=P^(m+2)+(P+1)^(2(m+1)-1)`
`=P^(m+2)+(P+1)^2(P+1)^(2m+1)`
`=P^(m+2)+(P+1)^2[k(P^2+P+1)-P^(m+1)]`
`=P^(m+2)+(P+1)^2.k(P^2+P+1)-(P+1)^2(P)^(m+1)`
`=P^(m+1)[P-(P+1)^2]+(P+1)^2.k(P^2+P+1)`
`=P^(m+1)[P-P^2-2P-1]+(P+1)^2.k(P^2+P+1)`
`=-P^(m+1)[P^2+P+1]+(P+1)^2,k(P^2+P+1)`
`=(P^2+P+1)[k.(P+1)^2-P^(m+1)]`
which is divisible by `P^2+P+1`, so the result is true for `n=m+1`. Therefore , the given result is true for all `n in N` by induction.