Sunday Afternoon Maths XIX Answers & Extensions

This post contains the answers to this week's Sunday Afternoon Maths and some extension problems based around the originals.

Dartboard

The shaded area is:

$$\pi (1)^2 - \pi (\frac{1}{2})^2 + \pi (\frac{1}{3})^2 - \pi (\frac{1}{4})^2 + \pi (\frac{1}{5})^2 - ...$$ $$=\sum_{i=1}^\infty \frac{\pi (-1)^{i-1}}{i^2}$$ $$=\pi\sum_{i=1}^\infty \frac{(-1)^{i-1}}{i^2}$$ $$=\pi\left(\frac{\pi^2}{12}\right)$$ $$=\frac{\pi^3}{12}$$

Extension

Prove that

$$=\sum_{i=1}^\infty \frac{(-1)^{i-1}}{i^2}=\frac{\pi^2}{12}$$

Multiples of Three

If \(10A+B=3n\), then:

$$100A+B = 100A-10A+10A+B$$$$=90A+3n$$$$=3(30A+n)$$

So \(A0B\div3=30A+n\).

Extension

What are \(A00B\div3\), \(A000B\div3\), \(A0000B\div3\), etc?