Question

For the given statement Pn, write the statements P1, Pk, and Pk+1.
2 + 4 + 6 + . . . + 2n = n(n+1)

Answer 1

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

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

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

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

Answer 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)