Get sum to a closed form ideas on how to start

Question

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.

Answer 1

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 $

Answer 2

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.