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Sigma Notation

Date: 09/07/2001 at 09:43:46
From: Steve Dockstader
Subject: Sigma Notation

Trying to prove that sigma (i^2) from i = 1 to n i equal to             
(n(n+1)(2n+1))/6 we were told to start with (i+1)^3 - i^3... 

I tried expanding this but I don't understand where that comes from 
in relation to i^2 and how it works.


Date: 09/07/2001 at 12:34:18
From: Doctor Peterson
Subject: Re: Sigma Notation

Hi, Steve.

Well, let's do it:

    (i+1)^3 - i^3 = i^3 + 3i^2 + 3i + 1 - i^3
                  = 3i^2 + 3i + 1

The difference on the left makes me think of telescoping sums: if we 
add up

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

the subtraction in one term cancels the addition in the previous term, 
making the sum

    (n+1)^3 - 1^3

So let's sum both sides of our expansion; I'll use sigma notation, 
which is indeed handy here, and will assume all sums go from 1 to n to 
save writing:

    SUM[(i+1)^3 - i^3) = SUM[3i^2 + 3i + 1]

    (n+1)^3 - 1 = 3 SUM[i^2] + 3 SUM[i] + SUM[1]

See if you can take it from here. After you're done, take a look here, 
where you'll see the same thing done without the benefit of sigma 

   Formula For the Sum Of the First N Squares   

Every method I know of for proving this formula requires some leap of 
insight, whether it's just knowing the formula to begin with and 
proving it's right, or doing some weird thing like this with seemingly 
unrelated sums.

Let me know if you need any more help finishing this.

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

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