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Sum of Consecutive Squares

Date: 05/11/2001 at 19:59:27
From: Robert DiPietro
Subject: Sum notation


I have searched over and over for how this equation was derived:

     Sum [i^2] = ((n)(n+1)(2n+1))/6


Thank you.

Date: 05/11/2001 at 20:12:35
From: Doctor Schwa
Subject: Re: Sum notation

Hi Robert,

My favorite proof goes like this.

Start with the sum of i^3 instead, for reasons that will become clear 
in a bit.

     1^3 + 2^3 + 3^3 + ... + n^3

Now let's compare that with:

     0^3 + 1^3 + 2^3 + ... + (n-1)^3

On the one hand, if you subtract the equal numbers, you have n^3 and 
all the rest of the numbers drop out.

On the other hand, if you subtract vertically, you have

   (1^3 - 0^3) + (2^3 - 1^3) + ... + (n^3 - (n-1)^3),

or in other words, sum from i = 1 to n of (i^3 - (i-1)^3). So that sum 
equals n^3.

Multiplying that out, it's the same as summing from i = 1 to n of 
(3i^2 - 3i + 1), which still equals n^3.

That is:

   3 * Sum [i^2] - 3 * Sum [i] + Sum [1] = n^3
   3 * Sum [i^2] - 3 * (n)(n+1)/2 + n    = n^3
   3 * Sum [i^2]                         = n^3 + 3n(n+1)/2 - n
       Sum [i^2]                         = (1/3) [n^3 + 3n(n+1)/2 - n]

Simplifying the expression on the right side gives you just what you 
were looking for.

This trick works in general: if you know how to sum 1, i and i^2, you 
can use it to figure out how to sum i^3, and so on.

There are lots of other methods, though, and you might find some of 
them by searching our Ask Dr. Math archives at:

   Search Dr. Math   

I found one explanation that gives a slightly different version of the 

   Formula For the Sum Of the First N Squares   

Two different approaches, finite differences and undetermined 
coefficients you might call them, are at:

   Finding sum formula using sequences of differences   


- Doctor Schwa, The Math Forum   
Associated Topics:
High School Sequences, Series

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