Get sum to a closed form ideas on how to start

I have this sum $1^2+ 2^2 + 3^2 + \ldots + x^2$

I started getting some sums to their closed forms. But I see that sometimes I start off on the bad track. So I'd like if it's possible some tips on how to start working on these not just some formulas.

For instance for this particular case I tried writing the term considering the next term using the formula $(x-1)^2 = x^2 -2x + 1$

So I got to $S=1^2 - 2*1+1 + 2^2-2*2+1+\ldots + n^2$ but then you get $S=S+\ldots$ you reduce S so you basically didn't accomplish anything. I often find myself starting off like this.

Is there a way to avoid this? What's the proper way to start this kind of problems?

I hope you get what I mean, I'll try to edit otherwise to be more clear. Thanks.

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Actually you accomplish something when you get that $S = S+\cdots$. You get a formula for $1+2+3+\cdots+n$. Take a hint from this and fiddle with a sum of power three and to get the one you are currently looking for. – NeverBeenHere Jan 8 '13 at 15:21
I find that for particually hard summations, it sometimes can be wise to form a difference equation of the sum $S_n$ - have you done difference equations/second order differential equations before? – Andrew D Jan 8 '13 at 16:13
@AndrewD I did in high-school/college but now I tried some problems that wouldn't have taken me more than 5 minutes to resolve and realized it takes me a while to figure out how to solve them without looking for formulas, so I just thought of some random problems to solve just for me, with no particular interest other than "fun". – Fofole Jan 9 '13 at 8:10

Let $$S_m=\sum_{1\le r\le n}r^m$$

Using the identity $$(r+1)^3-r^3=3r^2+3r+1$$

Putting $r=1,2,3,.\cdots,n$

$$(2)^3-1^3=3\cdot1^2+3\cdot1+1$$

$$(3)^3-2^3=3\cdot2^2+3\cdot2+1$$

** $$(n)^3-(n-1)^3=3(n-1)^2+3(n-1)+1$$

$$(n+1)^3-n^3=3n^2+3n+1$$

Adding them we get, $$(n+1)^3-1=3\sum_{1\le r\le n}r^2+3\sum_{1\le r\le n}r+\sum_{1\le r\le n}1=3S_2+3S_1+S_0$$

We know, $S_0=n$ and $S_1=\frac{n(n+1)}2$

So, $$3S_2=(n+1)^3-1-3\frac{n(n+1)}2-n=\frac{n(n+1)(2n+1)}2$$

In general to find $S_u=\sum_{1\le r\le n}r^u,$ we need to utilize the identity $(r+1)^{u+1}-r^{u+1}=\sum_{0\le t\le u}\binom u t r^t$ and put the values of $S_t$ for $0\le t\le u-1$

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It's not particularly formal, but if you're only looking at the form $\sum_{i = 0}^n i^p$ for some $p$, then the closed form should be a polynomial one degree higher. So for the example you've provided, you know it is $An^3 + Bn^2 + Cn + D$. Putting in four values of $n$ will give you a system of four linear equations, and solving will give you the coefficients. Once you find it, you can prove this "guess" with induction.

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