Verified Answer :

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\(P(n):2+4+6+\cdots+2n=\displaystyle\sum_{i=1}^n2i\)

\(P(1):\displaystyle\sum_{i=1}^12i=2=1(1+1)=2\)

\(P(k):\displaystyle\sum_{i=1}^k2i=2+\cdots+2k=k(k+1)\)

\(P(k+1):\displaystyle\sum_{i=1}^{k+1}2i=2+\cdots+2k+2(k+1)=(k+1)(k+2)\)

Answer:

\(P_{1}\) =  2

\(P_{k}\) = k(k+1)

\(P_{k+1}\) = (k+1)(k+2)

Step-by-step explanation:

We are given the statement,

\(P_{n}\) as 2 + 4 + 6 + . . . + 2n = n(n+1)

That is,

\(P_{n}\) as 2 + 4 + 6 + . . . + 2n = \(\sum_{i=1}^{n}2i\)

So, we have,

\(P_{1}\) = \(\sum_{i=1}^{1}2i\) = 2

\(P_{k}\) = \(\sum_{i=1}^{k}2i\) = 2 + 4 + 6 + . . . + 2k = k(k+1)

\(P_{k+1}\) = \(\sum_{i=1}^{k}2i\) = 2 + 4 + 6 + . . . + 2k + 2(k+1) = (k+1)(k+2)

Thus, we get,

\(P_{1}\) =  2

\(P_{k}\) = k(k+1)

\(P_{k+1}\) = (k+1)(k+2)

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