Finite Discrete difference calculus

[Johnny5]

Does anyone know how to use the finite discrete difference calculus to find a formula for the following finite sum?

 

[math] \sum_{n=1}^{n=m} n^2 [/math]

 

Thank you

[Dave]

The proof is quite nice; you have to use the fact that:

 

[math]\sum_{k=1}^{n} \left[(k+1)^3-k^3\right] = (n+1)^3-1[/math]

 

which is fairly obvious; just start writing it out and you'll see that all the terms cancel apart from the ones given.

 

Now also note that:

 

[math]\sum_{k=1}^{n} \left[(k+1)^3-k^3\right] = \sum_{k=1}^{n} (3k^2+3k+1)[/math]

 

By swapping things around and equating bits and pieces, you get:

 

[math](n+1)^3-1-n-\tfrac{3}{2}n(n+1) = 3\sum_{k=1}^{n} k^2[/math].

 

So by simplifying everything down:

 

[math]\sum_{k=1}^{n} k^2 = \frac{n(2n-1)(n+1)}{6}[/math].

[Johnny5]

The proof is quite nice; you have to use the fact that:

 

[math]\sum_{k=1}^{n} \left[(k+1)^3-k^3\right] = (n+1)^3-1[/math]

 

which is fairly obvious; just start writing it out and you'll see that all the terms cancel apart from the ones given.

 

Now also note that:

 

[math]\sum_{k=1}^{n} \left[(k+1)^3-k^3\right] = \sum_{k=1}^{n} (3k^2+3k+1)[/math]

 

By swapping things around and equating bits and pieces' date=' you get:

 

[math'](n+1)^3-1-n-\tfrac{3}{2}n(n+1) = 3\sum_{k=1}^{n} k^2[/math].

 

So by simplifying everything down:

 

[math]\sum_{k=1}^{n} k^2 = \frac{n(2n-1)(n+1)}{6}[/math].

 

This was terrific, thank you very much. I couldn't remember how to do it. There was some sort of procedure to start you off in the general case, which resembles the middle part of your proof. I'm still trying to recall it, I think it involved synthetic division. Hmm I'm sure it will come to me.

 

Thanks again