Sum of the sum of an arithmetic progression

a sequence is given as: 3,9,18,30,45....... (i will call this sequence A)

for this sequence the sum of the first 100 terms is needed

here is what i have done:

the difference of the above sequence is an arithmetic progression

ie: 3,6,9,12,15,........... (i will call this sequence B)

using some formula manipulation i have come up with an equation for the sequence 3,9,18,30,45,...................

which is

(n/4)*[(n+1)*(2a+(n-1)*d)+ (d/3) - ((n^2*d)/3)]

where n is the number of terms needed to be added

a is the first term of the sequence B (or sequence A since both first terms are the same)

d is the common difference in sequence B

using this formula i can come up with the answer but the formula is rather lengthy.....is there anyone who has a simpler formula or someone who can further simplify this one.

Thanks in advance

swnz