# Riemann Sum

1. Sep 14, 2008

### Brunll

About these formulas:

$$\frac{n(n+1)}{2}, \frac{n(n+1)(2n+1)}{6} ,(\frac{n(n+1)}{2})^2$$

How were they evolved?

And what about $$i^4, i^5, i^6,...$$

Thankyou everybody;;

2. Sep 14, 2008

### HallsofIvy

Staff Emeritus
Those are, of course, $\sum_{i=1}^n i$, $\sum_{i=1}^n i^2$, and $\sum_{i=1}^n i^3$, respectively. You can sum such things by using "Newton's Divided Difference" formula. It is a discrete version of Taylor's series. If f(n) is a function defined on the non-negative integers, $\Delta f$ is the "first difference", f(n)- f(n-1), $\Delta^2 f$ is the "second difference", $\Delta f(n)- \Delta f(n-1)$, $\Delta^3 f$ is the "third difference", etc. then
$$f(n)= f(0)+ \Delta f(0) n+ \frac{\Delta^2 f(0)}{2!} n(n-1)+ \frac{\Delta^3 f(0)}{3!} n(n-1)(n-2)+ \cdot\cdot\cdot[/itex] In particular, for a sum of terms in involving nk, $\Delta^{k+1} f$ is always 0 so this gives a polynomial of order k+1. A "short cut" to finding $\sum_{i=1}^n i^4$ is to set it equal to An5+ Bn4+ Cn3+ Dn2+ En+ F and use the first 6 values to get eqations for A, B, C, D, E and F. 3. Sep 14, 2008 ### Brunll Thakyou HallsofIvy for th explanation, but could you show me explicitly how to evolve he equation in [tex}i^4$$,for example.It would be very helpful.

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