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Find $S_r = 1^r + 2^r + 3^r + ... + n^r$ where r is any fixed positive integer in terms of $S_1, S_2, ... S_{r-1}$.

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What does Pythagoras' Theorem tell you about these angles: 90°, (45+x)° and (45-x)° in a triangle?

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Make a conjecture about the sum of the squares of the odd positive integers. Can you prove it?

Reciprocal Triangles

Stage: 5 Challenge Level: Challenge Level:2 Challenge Level:2

This solution was sent by Etienne from Parramatta Highschool, NSW Australia.

The $r$th triangular is $r(r+1)/2$ and it's reciprocal is $2/[r(r+1)]=2 \times [1/(r(r+1))]$

Now $1/r(r+1) = 1/r - 1/(r+1)$. Take note of this, very useful technique!

\[\frac{1}{r} - \frac{1}{r+1} = \frac{(r+1)-r}{r(r+1)} = \frac{1}{r(r+1)} \]

The sum of the reciprocals of the first n triangular numbers

\[\frac{2}{1 \times 2} + \frac{2}{2 \times 3} + \cdots + \frac{2}{n(n+1)} \] \[ = 2 \{ \frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \cdots + \frac{1}{n(n+1)} \} \] \[ = 2 \{ [\frac{1}{1} - \frac{1}{2}] + [\frac{1}{2} - \frac{1}{3}] + \cdots + [\frac{1}{n} - \frac{1}{(n+1)}] \} \]

Surprise, you get terms that cancel out each other, ie $-1/2$ and $1/2$, $-1/3$ and $1/3$, $-1/n$ and $1/n$. This is called 'telescoping'.

The sum thus equals \[ 2 \{ 1 - \frac{1}{(n+1)} \} \]

When n is large, $1/(n+1)$ is very small, so the sum is approximately $2$.

When $n$ tends to infinity, $1/(n+1)$ tends to $0$, and it turns out the infinite sum is exactly $2$.