# Geometric Sum (Power Series) Calculation

#### rwinston

Hi

In trying to calculate the following sum:

$$\sum_{i=1}^n{i^2}$$

I found the following expansions:

$$\sum_{i=1}^p \sum_{j=0}^{i-1} (-1)^j(i-j)^p {n+p-i+1\choose n-i} {p+1\choose j}$$

My question is: is there an easier or more intuitive way to compute the limit of the sum above?

#### Pere Callahan

Yes, it is infinity. And what you have is not a geometric series.

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#### tiny-tim

Science Advisor
Homework Helper
Hi rwinston! Hint: what is $$\sum_{i=1}^n{\left((i+1)^3\,-\,i^3}\right)$$ ? #### rwinston

Hi tim

So I think the expansion of that sum is

$$\sum_{i=1}^n{i^3+3i^2+3i+1} - \sum_{i=1}^n{i^3}$$
$$=\sum_{i=1}^{n}{3i^2+3i+1}$$
$$=3\sum_{i=1}^n{i^2}+3\sum_{i=1}^n{i}+n$$

Im not sure what the connection is yet tho...

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#### tiny-tim

Science Advisor
Homework Helper
Hi tim

So I think the expansion of that sum is

$$\sum_{i=1}^n{i^3+3i^2+3i+1} - \sum_{i=1}^n{i^3}$$
$$=\sum_{i=1}^{n}{3i^2+3i+1}$$
$$=3\sum_{i=1}^n{i^2}+3\sum_{i=1}^n{i}+n$$

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

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

So $$\sum_{i=1}^n{i^2}$$ is one-third the difference! ### Physics Forums Values

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