# Season 5 Episode 2

What's the difference of two squares? What's the difference of the difference of two squares? In **this episode** we're looking for patterns in sequences, and patterns across different sequences.

#### Further Reading

#### Binomial Expansion

$$(1+x)^3=1+3x+3x^2+x^3$$

Some people in the live chat seemed surprised that we could multiply out $(1+x)^3$ so easily. This is partly experience, and partly a bit of theory called the Binomial Theorem. In fact, there’s a link between multiplying out brackets and Pascal’s triangle (this is the Pascal’s triangle season, it seems!)

Find out more at Wikipedia | Binomial Theorem.

#### Telescoping Sums

These got a mention in chat, and they’re really cool. The idea is that if we’re adding together a sum like

$$(a-b)+(b-c)+...+(x-y)+(y-z)$$

then we can simplify that to $(a-z)$. Pretty obvious in that example, but here’s a better illustration of the technique; suppose we want to add together all the values of $(n^2-n)^{-1}$ from $n=2$ to $n=6$. Writing it out in full, that’s

$$\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}.$$

It’s not obvious how this could telescope down like the previous example. Here’s the trick; we can use the identity

$$\frac{1}{n^2-n}=\frac{1}{n-1}-\frac{1}{n}$$

on each term. Now our sum becomes

$$\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right) +\left(\frac{1}{4}-\frac{1}{5}\right) +\left(\frac{1}{5}-\frac{1}{6}\right),$$

which collapses down to $1-\frac{1}{6}$.

Sometimes this is handy when we want to calculate a sum “to infinity” of all the terms of a sequence; here we can deduce that

$$\sum_{n=2}^{N}\frac{1}{n^2-n}=1-\frac{1}{N}$$

and so

$$\sum_{n=2}^{\infty}\frac{1}{n^2-n}=1.$$

(Why “telescoping”? The thing to imagine is one of those old metal telescopes with lots of sections that fit inside each other so that it can collapse down small.)

There’s a similar calculation in a solution for a geometry problem hosted at AOPS Wiki | 2016 AMC 12B Problem 21

**Pi squared over six**

At the end of the livestream, I told you a few results with no proof at all. One of them was the eye-catching fact that

$$ \sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$$

What's that sum got to do with $\pi$? Where are the circles? I can't do better than link you to this Numberphile video (19 minutes, uses geometry to find the value of the sum).

For a more general method that can also be used for the $n^{-4}$ sum, see S3ep8 of the OOMC.

#### Why do we care about adding powers?

See the further reading from S3ep3 for an application of sums of powers to dice-rolling. That has an application to Dungeons & Dragons, of all things.

#### Why do we care about Dungeons & Dragons?

I just think it's neat.

#### Discrete calculus

Thanks Miles for the link to the video on the Mathologer channel (47 minutes of discrete calculus, includes something called “the Master formula”)

If you want to get in touch with us about any of the mathematics in the video or the further reading, feel free to email us on oomc [at] maths.ox.ac.uk