When P is a natural number,then {P^{n + 1}} + | vedclass

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(C) For $n = 1$,we get,${P^{1 + 1}} + {(P + 1)^{2(1) - 1}} = {P^2} + {(P + 1)^1} = {P^2} + P + 1$. This expression is clearly divisible by ${P^2} + P

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${P^2} + P + 1$

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(C) For $n = 1$,we get,${P^{1 + 1}} + {(P + 1)^{2(1) - 1}} = {P^2} + {(P + 1)^1} = {P^2} + P + 1$. This expression is clearly divisible by ${P^2} + P + 1$. Let us assume that the given result is true for $n = m \in N$,i.e.,${P^{m + 1}} + {(P + 1)^{2m - 1}} = k({P^2} + P + 1)$ for some $k \in N$ .....$(i)$ Now,for $n = m + 1$: ${P^{(m + 1) + 1}} + {(P + 1)^{2(m + 1) - 1}} = {P^{m + 2}} + {(P + 1)^{2m + 1}} = {P^{m + 2}} + {(P + 1)^2}{(P + 1)^{2m - 1}}$. Using equation $(i)$,we substitute ${(P + 1)^{2m - 1}} = k({P^2} + P + 1) - {P^{m + 1}}$: $= {P^{m + 2}} + {(P + 1)^2}[k({P^2} + P + 1) - {P^{m + 1}}] = {P^{m + 2}} + {(P + 1)^2} \cdot k({P^2} + P + 1) - {(P + 1)^2}{P^{m + 1}} = {P^{m + 1}}[P - {(P + 1)^2}] + {(P + 1)^2} \cdot k({P^2} + P + 1) = {P^{m + 1}}[P - ({P^2} + 2P + 1)] + {(P + 1)^2} \cdot k({P^2} + P + 1) = {P^{m + 1}}[-{P^2} - P - 1] + {(P + 1)^2} \cdot k({P^2} + P + 1) = -{P^{m + 1}}({P^2} + P + 1) + {(P + 1)^2} \cdot k({P^2} + P + 1) = ({P^2} + P + 1)[k{(P + 1)^2} - {P^{m + 1}}]$. Since this is a multiple of ${P^2} + P + 1$,the result is true for $n = m + 1$. By the principle of mathematical induction,the result is true for all $n \in N$.

When $P$ is a natural number,then ${P^{n + 1}} + {(P + 1)^{2n - 1}}$ is divisible by

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