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. Steve
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 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 notation: Formula For the Sum Of the First N Squares http://mathforum.org/dr.math/problems/sandin2.20.98.html 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 http://mathforum.org/dr.math/
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