Geometric Sum (Power Series) Calculation

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Discussion Overview

The discussion revolves around the calculation of the sum \(\sum_{i=1}^n{i^2}\), exploring various methods and expansions related to this series. Participants are examining both intuitive approaches and mathematical identities, with a focus on the geometric sum and power series concepts.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents a complex expansion for the sum \(\sum_{i=1}^n{i^2}\) involving nested sums and binomial coefficients.
  • Another participant asserts that the limit of the sum is infinity and states that the original expression is not a geometric series.
  • A different participant hints at a relationship involving the difference of cubes, suggesting a potential simplification.
  • One participant proposes an expansion of the sum that leads to a formulation involving \(3\sum_{i=1}^n{i^2}\) and \(3\sum_{i=1}^n{i}\), indicating a connection to the original sum.
  • There is a suggestion that \(\sum_{i=1}^n{i^2}\) is related to one-third of a difference, although the exact connection remains unclear to some participants.

Areas of Agreement / Disagreement

Participants express differing views on the nature of the series and the methods for calculating the sum. There is no consensus on a single, simpler method for computing \(\sum_{i=1}^n{i^2}\), and the discussion remains unresolved regarding the best approach.

Contextual Notes

Some participants reference mathematical identities and expansions without fully resolving the implications or assumptions behind them. The discussion includes various interpretations of the sum and its properties, which may depend on specific definitions or contexts.

rwinston
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Hi

In trying to calculate the following sum:

<br /> \sum_{i=1}^n{i^2}<br />

I found the following expansions:

<br /> \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}<br />

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

And what you have is not a geometric series.
 
Last edited:
Hi rwinston! :smile:

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

So I think the expansion of that sum is

<br /> \sum_{i=1}^n{i^3+3i^2+3i+1} - \sum_{i=1}^n{i^3}<br />
<br /> =\sum_{i=1}^{n}{3i^2+3i+1}<br />
<br /> =3\sum_{i=1}^n{i^2}+3\sum_{i=1}^n{i}+n<br />

Im not sure what the connection is yet tho...
 
Last edited:
rwinston said:
Hi tim

So I think the expansion of that sum is

<br /> \sum_{i=1}^n{i^3+3i^2+3i+1} - \sum_{i=1}^n{i^3}<br />
<br /> =\sum_{i=1}^{n}{3i^2+3i+1}<br />
<br /> =3\sum_{i=1}^n{i^2}+3\sum_{i=1}^n{i}+n<br />

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! :smile:
 

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