Sum of the sum of an arithmetic progression

**swanz** · September 25th 2009, 06:08 PM

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

**red_dog** · September 26th 2009, 12:00 AM

$a_1=3=3\cdot 1$

$a_2=9=3(1+2)$

$a_3=18=3(1+2+3)$

$a_4=30=3(1+2+3+4)$

$a_5=45=3(1+2+3+4+5)$

Then $a_n=3(1+2+3+\ldots+n)=3\cdot\frac{n(n+1)}{2}=\frac {3}{2}(n^2+n)$

$\sum_{k=1}^{100}a_k=\frac{3}{2}\sum_{k=1}^{100}(k^ 2+k)=\frac{3}{2}\left(\sum_{k=1}^{100}k^2+\sum_{k= 1}^{100}k\right)$

Now use the formula $\sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}{6}$ and $\sum_{k=1}^nk=\frac{n(n+1)}{2}$ replacing n with 100.

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