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Quick note: This is a short and hopefully fun text written by, and aimed at high school calculus student(s), with a focus on intuition. It covers two topics in a way that I've endeavoured to make engaging. All JS graphics are by me. Please enjoy!

Welcome to my blog. I am an A Level (High School) student studying Maths, Further Maths, Chemistry and Physics in the UK. On this site, I hope to share my love for maths and its sister disciplines through a range of different articles. I recommend that you read my introductory post at some point. Otherwise, I hope you enjoy my work! Comments, queries, corrections and complaints are always welcome to rory@chalkplainmaths.co.uk.

Previous to this, we made the first of the two puzzle pieces that we will soon slot together to form our proof. Prior to unification, however, we must create the second. Once again, we are dealing with two variables, however, rather than being the powers of the numerator and denominator, this time, they are both in the denominator, for we are going after the proof of our statement where there are two distinct factors beneath the fraction. Expressed in symbols, our aim is to prove that constant \(A,B\)s exist where \[ \frac{1}{(x+a)^n(x+b)^m}\equiv\sum_{r=1}^n\frac{A_r}{(x+a)^r}+\sum_{r=1}^m\frac{B_r}{(x+b)^r} \] \[ \textrm{where} \quad n,m\in\mathbb{N} \quad \textrm{and} \quad a\neq b \textrm{.} \] The \(\sum\) symbol means to sum together the expression inside for all values between those given on the top and bottom (inclusive). Note also that as \(n\) and \(m\) are natural numbers, they are nonzero, so both factors are to a power of at least one. Additionally, since \(a\neq b\) we know the factors are definitely different. This means we don't have to worry about edge cases where the form looks different to what is above.

In this chapter, we are going to tackle the first section of our proof, using a method known as 'Proof by Induction'. It allows us to induce the truth of a statement over all terms in a sequence having only proved that it holds for the first term, and that if it's true for any random term, it also holds for the next. Imagine a line of pillars out at sea, evenly spaced apart and stretching into the horizon. If I place you on the first pillar, and build you a portable bridge that spans the gap between one pillar and the next, it follows that you will be able to reach any pillar, even if the line is infinitely long!