Question

Sk+1=Sk+ak+1=(6+12+18+24+...+6k)+ak+1. ak+1=

Answer 1

I'm guessing you mean

`S_(k+1) = S_k + a_(k+1)`

and that
`S_k` is the sum

`S_k = 6 + 12 + 18 + \ldots + 6k = 6 (1 + 2 + 3 + \ldots + k) = 3 k (k+1)`

using the well-known formula

`\displaystyle \sum_(i=1)^n i = 1 + 2 + 3 + \cdots + n = \frac{n(n+1)}2`

Then by substitution,

`S_(k+1) = 6 (1 + 2 + 3 + \cdots + k + (k+1)) = 3 (k+1) (k+2)`

and hence

`a_(k+1) = S_(k+1) - S_k`

`a_(k+1) = 3 (k+1) (k+2) - 3k (k+1)`

`a_(k+1) = 3 (k+1) \bigg((k+2) - k\bigg)`

`\implies \boxed{a_(k+1) = 6 (k + 1)}`