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### Sigma Notation

```
Date: 09/07/2001 at 09:43:46
Subject: Sigma Notation

Trying to prove that sigma (i^2) from i = 1 to n i equal to

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

(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
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/
```
Associated Topics:
High School Sequences, Series

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