Finding sum of an exponential (not geometric) series

[alkatran]

If I want to find the sum of a series that is = x^2, how can I do it without adding each individual element?

For example, with linear series, you take the lowest and highest values (at the beginning and end) to get the average, then multiply by the number of elements. I've been trying to figure out one for

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

but keep hitting dead ends. Care to help?

[bugzpodder]

n(n+1)(2n+1)/6

[alkatran]

Could you explain how you found it? Just searched? I worked for awhile trying to figure it out on my own.

[sql_lall]

the patterns can be found like this:

terms:
1 5 14 30 55
differences:
4 9 16 25
again:
5 7 9
again:
2 2
again:
0

u see how it took 5 steps? 5-2 = 3, so it's a cubic solution

if it takes n steps, it's a solution of degree n-2.

just try ax^3 + bx^2 + c^x + d

[alkatran]

You know I was basing my entire problem on that regularity? (the +1, +3, +5) I just didn't consider actually shoving it into an equation!

I hope this doesn't mean geometric series can't be summed... (since their pattern just moves to the right if you try adding, only multiplying works..)

[sql_lall]

Sure they can:
x^1 + x^2 + x^3 + .... + x^n = [x^(n+1) - 1]/[x-1]

All these things are part of a general technique called 'differential equations'...sounds like calculus, i know.


Anyway, the series above is basically:

S₁ = x;
S_n = S_n-1 + x^n

there are some tricks to working these out, basically which give the formula above.