\(n+2\) terms, since to get 1 using the formula \(6n+7\) we must use \(n=-1\text{.}\) Thus we have \(n\) terms, plus two, when \(n=0\) and \(n=-1\text{.}\)

\(6n+1\text{,}\) which is 6 less than \(6n+7\) (or plug in \(n-1\) for \(n\)).

\(\frac{(6n+8)(n+2)}{2}\text{.}\) Reverse and add. Each sum gives the constant \(6n+8\) and there are \(n+2\) terms.