In the given series, there is sum of multiple of corresponding terms of three A.P.’s The three A.P.’s are
(i) 1, 2, 3, 4, ..............
(ii) 2, 3, 4, 5, ..............
(iii) 3, 4, 5, 6,...............
Now, the nth term of sum is
an = (nth term of 1st A.P.) × (nth term of 2nd A.P.) × (nth term of 3rd A.P.)
= n(n + 1) (n + 2) = n3+3n2 + 2n
Hence, the sum of n terms is Sn =

n

Σ

k=1

ak
⇒ Sn =

n

Σ

k=1

(k3+3k2+2k) =

n

Σ

k=1

k3+3

n

Σ

k=1

k2+2

n

Σ

k=1

k
= [

n(n+1)

2

]2 + 3[

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

6

] + 2[

n(n+1)

2

]
=

n(n+1)

2

[

n(n+1)

2

+(2n+1)+2] =

n(n+1)

2

[

n2+n+4n+2+4

2

]
=

n(n+1)(n2+5n+6)

4

=

n(n+1)(n+2)(n+3)

4