Find Closed-Form Expression for Summation of n^2

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Discussion Overview

The discussion revolves around finding a closed-form expression for the summation of the squares of the first N positive integers, represented as Q_N = ∑_{z=1}^N z^2. The scope includes mathematical reasoning and exploration of different approaches to derive the expression.

Discussion Character

  • Mathematical reasoning
  • Exploratory
  • Debate/contested

Main Points Raised

  • One participant requests hints for finding a closed-form expression for Q_N.
  • Another participant proposes the solution N(N+1)(2N+1)/6 and suggests using the identity (k+1)^3 - k^3 = 3k^2 + 3k + 1, summing it from k=1 to N.
  • A different participant questions the utility of the proposed approach, noting that it requires evaluating a summation of k^2, which is the original problem.
  • One participant asserts that the leftmost summation is a telescopic sum that equals (N+1)^3 - 1, leading to an equation for Q_N.
  • A later reply acknowledges the telescopic nature of the sum and expresses realization about the oversight.
  • Another participant reiterates the initial question and introduces an alternative equation form: (n^3 - n)/24, noting that sums of squares are a special case and mentioning specific integers for which the sum of squares is also a square.

Areas of Agreement / Disagreement

Participants express differing views on the proposed methods for finding the closed-form expression, with no consensus reached on the best approach or the validity of the various claims made.

Contextual Notes

Some participants' arguments depend on the assumptions about the properties of telescopic sums and the evaluation of summations, which remain unresolved. The discussion also touches on the relationship between sums of squares and specific integer values.

mnb96
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Hello,
could anyone give me a hint on how to find a closed-form expression for the following summation:

[tex]Q_N = \sum_{z=1}^N z^2[/tex]

where z are positive integers from 1 to N.
Thanks.
 
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The solution is N(N+1)(2N+1)/6. The trick is simple: take the equations: (k+1)^3-k^3=3k^2+3k+1, sum them up from k=1 to N.
 
csopi said:
The solution is N(N+1)(2N+1)/6. The trick is simple: take the equations: (k+1)^3-k^3=3k^2+3k+1, sum them up from k=1 to N.

Uhm...you mean:

[tex]\sum_{k=1}^N \left( (k+1)^3-k^3 \right) = \sum_{k=1}^N \left( 3k^2+3k+1\right)[/tex]

But how can this help, if we don't know the value of the leftmost summation?
Evaluating the rightmost term also requires evaluating a summation of k2, which is what we are trying to find :confused:
 
You do know, it's a telescopic sum and equals (N+1)^3-1. After that, you get an equation for Q_N
 
csopi said:
You do know, it's a telescopic sum and equals (N+1)^3-1.

Ups! that's true...I should have noticed it immediately. Thanks.
 
mnb96 said:
Hello,
could anyone give me a hint on how to find a closed-form expression for the following summation:

[tex]Q_N = \sum_{z=1}^N z^2[/tex]

where z are positive integers from 1 to N.
Thanks.

Also note the following equation form: (n^3 - n)/24. The sums of squares are a special case of this equation form, and incidentally, there are only 3 integers in N | the sum of squares is also square. This holds for 0, 1, 24 and no other.

- RF
 

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