In the given series
an = 1+22 + ... + n2 =

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

6

=

(n2+n)(2n+1)

6

=

3n3+2n2+n2+n

6

=

2n3+3n2+n

6

Hence, the sum to n terms is
Sn =

n

Σ

k=1

ak =

1

6

Σ k

k=1

(2k3+3k2+k) =

1

6

[2

n

Σ

k=1

k3+3

n

Σ

k=1

k2+

n

Σ

k=1

k]
=
=

n(n+1)

2×6

[n (n + 1) + (2n + 1) + 1] =

n(n+1)

12

[n2+n+2n+2]
=

n(n+1)

12

(n + 1) (n + 2) =

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

12