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April 16, 2014

April 16, 2014

Posted by **Sean** on Monday, May 30, 2011 at 10:33pm.

First, I solve analytically so I know the answer I am trying to reach: 4/3 * x^3 over [0,4] = 4/3 * 4^3 = 256/3

Now, by approximating sums, I can get:

256 * lim(n -> infinity) of sum(j=0 to n-1) of j^2/n^3

I can use the computer to solve this and I reach the correct answer. In Mathematica Alpha notation: 256 * lim sum j^2/n^3, j=0..n as n -> infinity

How do I solve that limit by hand? Or is there a better way to solve the original problem?

- Calculus -
**Count Iblis**, Tuesday, May 31, 2011 at 2:45pmYou need to derive the formula for:

Sum from j = 0 to N of j^2

There are many different ways to do this, the more advanced math you know, the simpler it gets :).

An elementary method is to consider summing (j+1)^3 - j^3 instead of j^2. Obviously, if you sum a function of the form f(j+1) - f(j), all the terms except the first and last one will cancel:

Sum from j = 0 to N of [f(j+1) - f(j)] =

f(N+1) - f(0)

If we choose f(j) = j^3, then:

f(j+1) - f(j) =

(j+1)^3 - j^3 =

3 j^2 + 3 j + 1

So, if you know the summation of j from zero to N, you can find the summation of j^2. Of course, you can find the formula for the summation of j in the same way by taking f(j) = j^2.

- Calculus -
**Sean**, Tuesday, May 31, 2011 at 3:44pmYou are awesome Iblis. That limit was the tricky part. I followed your proof and it makes perfect sense. Thanks

- Calculus -
**Sean**, Tuesday, May 31, 2011 at 3:45pmI meant that summation was the tricky part (not limit). thanks again.

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