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When dealing with the summation of large numbers of, well, numbers, it can become very tedious to write everything out, especially when you're trying to add up arbitrary numbers of things, such as with polynomials.

The solution to the problem that most people learn first is ellipsis notation, which can be quite clunky, but is very easy to read:

1, 2, 3, ..., *n*

obviously means "

Count one,

two,

three, and continue on until you have reached

*n*".

When more complicated sums come up though, this notation quickly becomes nearly impossible to understand. Say you want to sum a series in which there are constant and variables combined with all kinds of addition, subtraction, multiplication, division, derivation and integrations involved? Well, you'd want an easier, more obvious way of looking at things.

And so we have sigma notation. A capital letter sigma looks like this: Σ

This letter stands for "Sum"

A more formal definition goes as follows:<
blockquote>If a_{m}, a_{m+1}, a_{m+2}, ... a_{n} are real numbers and *m* and *n* are integers such that *m* ≤ *n*, then

*n*

Σ a_{i} = a_{m} + a_{m+1} + a_{m+2} + ... a_{n-1} + a_{n}

*i=m*

In this way, we can write out the sum of the first six digits as:

*6*

Σ *i*

*i=1*

Which, by the definition above, is equivalent to writing

1 + 2 + 3 + 4 + 5 + 6

The use of this should be instantly seen-- you can sum anything like this, and do so without actually writing down each

element, which would take up way too much space and take way too long anyway. But beyond that, a few

theorems have been discovered about sums which are even more useful. The first three should be pretty obvious:

Where *c* is any constant (not depending on *i*:

*n* n
Σ *ca*_{i} = *c*Σ *a*_{i}
*i=m i=m*
*n* n n
Σ *(a*_{i} ± *b*_{i}) = Σ *a*_{i} ± Σ *b*_{i}
*i=m i=m i=m*

The first of these two rules becomes clear when you write out the sigma notation as it is defined. Then you just apply the

distributive property of real numbers.

*ca*_{m} + ca_{m+1} + ... + ca_{n} = c(a_{m} + a_{m+1} + ... + a_{n})

The second rule is just an application of the

associative property and the

commutative property.

*(a*_{m} ± b_{m}) + (a_{m+1} ± b_{m+1}) + ... + (a_{n} ± b_{n}) = (a_{m} + a_{m+1} + ... + a_{n}) ± (b_{m} + b_{m+1} + ... + b_{n})

The third formula is nearly as simple.

*n*

Σ *1* = *n*

*i=1*

This is just a case of multiplying 1, n times. Since we all know that anything times 1 is that thing, the sum of n 1's is n.

Now it's time to get into the more interesting formulas.

*n*

Σ *i* = *n(n + 1) / 2*

*i=1*

This was proved by

Karl Friedrich Gauss,

apparently when he was ten years old. The proof is as follows:

Write the sum S twice, once in the usual order and once in the reverse order.
S = 1 + 2 + 3 + ... + (n-2) + (n-1) + n
S = n + (n-1) + (n-2) + ... + 3 + 2 + 1

Now, if you add the two sums together, you get:
2S = (n+1) + (n+1) + (n+1) + ... + (n+1) + (n+1) + (n+1)

Interesting, no? Every term is (n+1) all of a sudden. And not only that, but there are n terms, so now we can write:
2S = n(n+1)

Or:
S = n(n+1)/2

Which is what we were trying to prove! Next up we have the sums of squares--

*n*

Σ *i *^{2} = *n(n + 1)(2n+1) / 6*

*i=1*

Proof by

mathematical induction:

Let S_{n} be equal to the formula above.
S_{1} is true because

1^{2} = 1(1+1)(2*1+1)/6

Assume that S_{k} is true, which is to say that1^{2} + 2^{2} + 3^{2} + ... + k^{2} = k(k+1)(2k+1)/6

Then
1^{2} + 2^{2} + 3^{2} + ... + (k+1)^{2} = (1^{2} + 2^{2} + 3^{2} + ... + k^{2}) + (k + 1)^{2}
(Because we're assuming that S_{k} is true...)

= (k(k+1)(2k+1)/6) + (k+1)^{2}

= (k+1) (k(2k+1)+(6k+1)/6)

= (k+1) (2k^{2}+7k+6)/6)

= ( (k+1)(k+2)(2k+3) )/6

= ( (k+1){(k+1)+1}{2(k+1)+1} )/6

So S_{k+1} is true.

By the principle of mathematical induction, S_{n} is true for all n.

The following theorems I will not prove, but were taken from http://polysum.tripod.com/

*n*

Σ *i* ^{3} = {n(n + 1)/2}^{2}

*i=1*

*n*

Σ *i* ^{4} = n(2n+1)(n+1)(3n^{2}+3n-1)/30

*i=1*

*n*

Σ *i* ^{5} = (n^{2}(2n^{2}+2n-1)(n+1)^{2})/12

*i=1*

*n*

Σ *i* ^{6} = (n(2n+1)(n+1)(3n^{4}+6n^{3}-3n+1))/42

*i=1*

*n*

Σ *i* ^{7} = (n^{2}(3n^{4}+6n^{3}-n^{2}-4n+2)(n+1)^{2})/24

*i=1*

*n*

Σ *i* ^{8} = (n(2n+1)(n+1)(5n^{6}+15n^{5}+5n^{4}-15n^{3}-n^{2}+9n-3))/90

*i=1*

*n*

Σ *i* ^{9} = (n^{2}(n^{2}+n-1)(2n^{4}+4n^{3}-n^{2}-3n+3)(n+1)^{2})/20

*i=1*

*n*

Σ *i* ^{10} = (n(2n+1)(n+1)(n^{2}+n-1)(3n^{6}+9n^{5}+2n^{4}-11n^{3}+3n^{2}+10n-5))/66

*i=1*

Reference Used:

Stewart, James "Single Variable Calculus, Fifth Edition", Thomas Learning Acedemic Resource Center ©2003