Math Is Fun Forum

Alex Z.

Guest

Sum of sums of arithmetic progressions?

Imagine a series where each member is the sum of members of an arithmetic series, up to its position. For example:
b1=a1,
b2=a1+a2
b3=a1+a2+a3
bn=a1+...+an=n/2*(a1+an)
Is there a formula for summing the b series?

A little background; I got to this by trying to work out a formula for a sum of a series where each member is the square of its position (b1=1, b2=4, b3=9 etc.). I quickly found out that  the difference between one member and the one before is a simple arithmetic series (a1=b2-b1=3, a2=b3-b2=5, etc.) and that each member of the b series can be expressed as a sum of a:
a1=3, an=a1+2(n-1), bn=n²=1+∑an=1+(n-1)(a1+an-1)/2 for n>1.[2²=1+(2-1)(3+3)/2=4]
This means that ∑bn=1+∑a1+∑a2+∑a3+...∑an-1 and I just can't figure out how to simplify that.
Can anyone help me?

tnx

luca-deltodesco

Member

Registered: 2006-05-05

Posts: 1,470

Re: Sum of sums of arithmetic progressions?

where d = a[sub]n[/sub]-a[sub]n-1[/sub] the common difference

Last edited by luca-deltodesco (2008-12-11 22:54:35)

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Alex Z.

Guest

Re: Sum of sums of arithmetic progressions?

Thank you!