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

`P_(1)` = 2

`P_(k)` = k(k+1)

`P_(k+1)` = (k+1)(k+2)

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)