If P(n) = 2 + 4 + 6 + dots + 2n,n in N,then P | vedclass

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(D) The given statement $P(n) = 2 + 4 + 6 + \dots + 2n$ is an arithmetic series with sum $S_n = n(n + 1)$.Given the inductive step $P(k) = k(k + 1) +

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(D) The given statement $P(n) = 2 + 4 + 6 + \dots + 2n$ is an arithmetic series with sum $S_n = n(n + 1)$.Given the inductive step $P(k) = k(k + 1) + 2 \implies P(k + 1) = (k + 1)(k + 2) + 2$,we are testing the validity of the formula $P(n) = n(n + 1) + 2$.Since the actual sum of the first $n$ even numbers is $n(n + 1)$,the formula $P(n) = n(n + 1) + 2$ is incorrect for all $n \in N$.Therefore,the inductive step provided does not prove the formula for any $n$,and thus nothing can be concluded about the validity of the formula $P(n) = n(n + 1) + 2$ from the given information.

If $P(n) = 2 + 4 + 6 + \dots + 2n$,$n \in N$,then $P(k) = k(k + 1) + 2 \implies P(k + 1) = (k + 1)(k + 2) + 2$ for all $k \in N$. So we can conclude that $P(k) = k(k + 1) + 2$ for all $k \in N$ is true. What can we conclude about $P(n) = n(n + 1) + 2$ for all $n \in N$?

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