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A tower of squares is built inside a right angled isosceles
triangle. The largest square stands on the hypotenuse. What
fraction of the area of the triangle is covered by the series of
squares?

Find the sum of the series.

Weekly Problem 20-2013

# Sums of Powers - A Festive Story

##### Stage: 3 and 4

Article by Theo Drane

The general case:

\begin{eqnarray} 1^m +2^m + \dots + n^m \\ = \left.(1^me^t +
2^me^{2t} + \dots n^me^{nt})\right|_{t=0}\\ =\left.
\frac{d^m}{dt^m}\left(1 +e^t + e^{2t} + \dots + e^{nt}\right)
\right|_{t=0}\\ =\left. \frac{d^{m+1}}{dt^{m+1}}\frac{t(e^{(n+1)t}
- 1)}{e^t-1} \right|_{t=0}\\ =\frac{1}{m+1}\left( \left.
\frac{d^{m+1}}{dt^{m+1}}\frac{te^{(n+1)t}}{e^t-1}\right|_{t=0}
-\left.\frac{d^{m+1}}{dt^{m+1}}\frac{t}{e^t-1}\right|_{t=0}\right)
\\ =\frac{1}{m+1}\left( \left.
\frac{d^{m+1}}{dt^{m+1}}\sum_{k=0}^{\infty}B_k(n+1)\frac{t^k}{k!}\right|_{t=0}
-\left.\frac{d^{m+1}}{dt^{m+1}}\sum_{k=0}^{\infty}B_k\frac{t^k}{k!}\right|_{t=0}\right)
\\ =\frac{B_{m+1}(n+1) - B_{m+1}}{m+1}\\ \end{eqnarray}

Where $B_n(x)$ is the Bernoulli polynomial and $B_n$ are the
Bernoulli numbers

Conclude that $$1^m + 2^m + \dots + n^m = \frac{B_{m+1}(n+1) -
B_{m+1}}{m+1}$$ Amongst other things, all you have to do now is
find out what on earth is a Bernoulli polynomial!!