how do u get that?

I got it from 'Concrete Mathematics, A foundation for computer science', by Graham,Knuth,Patashnik.

This is a really cool book that goes in great depths into this stuff. I wish I had taken a class in this to learn it better, but oh well.

Section 2.5 covers several different methods for evaluating this exact sum.

Method 0 : Look it up. This is the method I used in answering your question at first.

Method 1 : Guess the answer, then prove it. Not much more interesting.

Method 2 is clever and nifty, however:

Consider the sum of the *cubes* from 1 to N, which I'll call Cn since I'm too lazy to put in the subscript codes.

Cn = sum(k=1..n, k

^{3}) // note more laziness with notation

I'll also define

Sn = sum(k=1..n, k

^{2}) which is what you want to find.

Also note that sum(k=1..n, k) = (1/2)(n)(n+1), I assume you've already figured that one out.

then...

Cn+1 = Cn + (n+1)

^{3} = sum(k=1..n+1, k

^{3})

Next change variables on the sum, to l where l = k-1, and so

Cn + (n+1)

^{3} = sum(l=0..n, (l+1)

^{3})

= sum(l=0..n, l

^{3} + 3*l

^{2} + 3*l + 1)

= 1 + sum(l=1..n, l

^{3} + 3*l

^{2} + 3*l + 1)

= 1 + sum(l=1..n, l

^{3}) + 3 * sum(l=1..n, l

^{2}) + 3 * sum(l=1..n, l) + sum(l=1..n, 1)

= 1 + Cn + 3 * Sn + 3 * (1/2)(n)(n+1) + n

The Cn can be subtracted off both sides of this equation, thus removing it. This is convenient since we didn't actually care about it in the first place.

Rearrange a bit, to solve for Sn which is what you want,

Sn = (1/3)*((n+1)

^{3} - 1 - (3/2)(n)(n+1) - n)

a little more algebra then beats this into the final form

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