In the given series, there is a sum of multiple of corresponding terms of two A.P.’s.
The two A.P.’s are
(i) 1, 2, 3, 4, .......
(ii) 2, 3, 4, 5, .....
Now the nth term of sum is
an = (nth term of first A.P.) × (nth term of second A.P.)
= n × (n + 1) = n2 + n
Since, the sum to n terms is Sn =

n

Σ

k=1

ak
⇒ Sn =

n

Σ

k=1

(k2+k) =

n

Σ

k=1

k2+

n

Σ

k=1

k =

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

6

+

n(n+1)

2

=

n(n+1)

2

[

2n+1

3

+1] =

n(n+1)(2n+4)

6

=

n(n+1)(n+2)

3