If $P(n) = 2 + 4 + 6 + ... + 2n,\ n \in N,\ then\ P(k) = k(k + 1) + 2 \Rightarrow P(k + 1) = (k + 1)(k + 2) + 2\ for\ all\ k \in N$. So, we can conclude that $P(n) = n(n + 1) + 2$ for

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Dr. Abubacker · Accepted Answer

(d)
It is obvious, nothing can be said.

Q. If $P(n) = 2 + 4 + 6 + ... + 2n,\ n \in N,\ then\ P(k) = k(k + 1) + 2 \Rightarrow P(k + 1) = (k + 1)(k + 2) + 2\ for\ all\ k \in N$. So, we can conclude that $P(n) = n(n + 1) + 2$ for

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