Let $S_n$ = 3 + 7 + 13 + 21 + 31 + ..... + $a_{n-1} + a_n$ ....(i) or $S_n$ = 3 + 7 + 13 + 21 + ...... + an – 2 + $a_{n-1} + a_n$ ....(ii) On subtracting (ii) from (i), we get 0 = 3 + [4 + 6 + 8 + ..... (n – 1) terms] – $a_n$ ⇒ $a_n$ = 3 + $\frac{n-1}{2}$ [8 + (n - 2) . 2] Since $S_n$ = $\frac{n}{2}$ [2a + (n - 1) d] ⇒ $a_n$ = 3 + (n - 1) (n + 2) ⇒ $a_n$ = 3 + $n^2$ + n - 2 ⇒ $a_n$ = $n^2$ + n + 1 Hence $S_n$ = $\sum\limits_{k=1}^{n} a_k$ = $\sum\limits_{k=1}^{n} (k^2+k+1)$ = $\sum\limits_{k=1}^{n} k^2 + \sum\limits_{k=1}^{n} k + \sum\limits_{k=1}^{n} 1$ = $\frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}$ + n = $\frac{n}{6}$ [(n + 1) (2n + 1) + 3 (n - 1) + 6] = $\frac{n}{6}$ [$2n2$ + n + 2n + 1 + 3n + 3 + 6] = $\frac{n}{6}$ [$2n^2$ + 6n + 10] = $\frac{n}{3}$ ($n^2$ + 3n + 5)