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ThePerfectHacker · #4 February 24th, 2008, 01:51 PM

2)We are given that $\sum_{k=1}^n k^m = P_m(n)$ where $P_m$ is a polynomial. Suppose we wish to compute the integral of $f(x) = x^m$ over the interval $[0,1]$ . Then $\int_0^1 x^m ~ dx = \frac{1}{m+1}$ . Now there is another way to compute this integral, and that is to use a Riemann sum with $n$ equal subdivision point using the right endpoint. In that case we get $\lim ~ \left( \sum_{k=1}^n \frac{k^m}{n^m} \right) \cdot \frac{1}{n} = \lim ~ \frac{1}{n^{m+1}} \sum_{k=1}^n k^m = \lim ~ \frac{P_m(n)}{n^{m+1}}$ . This limit needs to be $1/(m+1)$ since that is the integral. But this means the polynomial $P_m(n)$ must be degree $m+1$ with leading coefficient of $1/(m+1)$ .

In fact, note the following,
$\sum_{k=1}^n k = \frac{n(n+1)}{2} = \boxed{\frac{1}{2}}n^2 + \frac{1}{2}n$ .
$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6} = \boxed{\frac{1}{3}}n^3 + \frac{1}{2}n^2+\frac{1}{6}n$ .
$\sum_{k=1}^n k^3 = \frac{n^2(n+1)^2}{4} = \boxed{\frac{1}{4}}n^4 + \frac{1}{2}n^3+\frac{1}{4}n^2$ .
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