Let Sn = 3 + 7 + 13 + 21 + 31 + ..... + an–1+an ....(i)
or Sn = 3 + 7 + 13 + 21 + ...... + an – 2 + an–1+an ....(ii)
On subtracting (ii) from (i), we get
0 = 3 + [4 + 6 + 8 + ..... (n – 1) terms] – an
⇒ an = 3 +

n−1

2

[8 + (n - 2) . 2]
Since Sn =

n

2

[2a + (n - 1) d]
⇒ an = 3 + (n - 1) (n + 2)
⇒ an = 3 + n2 + n - 2 ⇒ an = n2 + n + 1
Hence Sn =

n

Σ

k=1

ak =

n

Σ

k=1

(k2+k+1) =

n

Σ

k=1

k2+

n

Σ

k=1

k+

n

Σ

k=1

1
=

n(n+1)(2n+1)

6

+

n(n+1)

2

+ n =

n

6

[(n + 1) (2n + 1) + 3 (n - 1) + 6]
=

n

6

[2n2 + n + 2n + 1 + 3n + 3 + 6] =

n

6

[2n2 + 6n + 10]
=

n

3

(n2 + 3n + 5)