Geometric Sum (Power Series) Calculation

  • Thread starter rwinston
  • Start date
Hi

In trying to calculate the following sum:

[tex]
\sum_{i=1}^n{i^2}
[/tex]

I found the following expansions:

[tex]
\sum_{i=1}^p \sum_{j=0}^{i-1} (-1)^j(i-j)^p {n+p-i+1\choose n-i} {p+1\choose j}
[/tex]

My question is: is there an easier or more intuitive way to compute the limit of the sum above?
 
Yes, it is infinity. :smile:

And what you have is not a geometric series.
 
Last edited:

tiny-tim

Science Advisor
Homework Helper
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Hi rwinston! :smile:

Hint: what is [tex]\sum_{i=1}^n{\left((i+1)^3\,-\,i^3}\right)[/tex] ? :smile:
 
Hi tim

So I think the expansion of that sum is

[tex]
\sum_{i=1}^n{i^3+3i^2+3i+1} - \sum_{i=1}^n{i^3}
[/tex]
[tex]
=\sum_{i=1}^{n}{3i^2+3i+1}
[/tex]
[tex]
=3\sum_{i=1}^n{i^2}+3\sum_{i=1}^n{i}+n
[/tex]

Im not sure what the connection is yet tho...
 
Last edited:

tiny-tim

Science Advisor
Homework Helper
25,790
242
Hi tim

So I think the expansion of that sum is

[tex]
\sum_{i=1}^n{i^3+3i^2+3i+1} - \sum_{i=1}^n{i^3}
[/tex]
[tex]
=\sum_{i=1}^{n}{3i^2+3i+1}
[/tex]
[tex]
=3\sum_{i=1}^n{i^2}+3\sum_{i=1}^n{i}+n
[/tex]

Im not sure what the connection is yet tho...
ah… but you know what [tex]3\sum_{i=1}^n{i}\,+\,n[/tex] is;

And you also know what [tex]\sum_{i=1}^n{\left((i+1)^3\,-\,i^3}\right)[/tex] is … don't you … ?

So [tex]\sum_{i=1}^n{i^2}[/tex] is one-third the difference! :smile:
 

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