# 3.5 The binomial theorem

## 3.5.1 Derivation

Note that this definitely isn’t on any A Level specification.

A binomial is something in the form

$(a+b)^{n}$

What we’re interested in is how to expand this bracket for large values of $n$. Small values of $n$ are not too bad, e.g. $(a+b)^{2}$ 55 5 If you’ve no idea how to expand this, review a GCSE textbook. can be expanded like this

 $\displaystyle(a+b)^{2}$ $\displaystyle=(a+b)(a+b)$ $\displaystyle=(a+b)a+(a+b)b$ $\displaystyle=a^{2}+ab+ab+b^{2}$ $\displaystyle=a^{2}+2ab+b^{2}$ (3.56)

What about $(a+b)^{3}$?

 $\displaystyle(a+b)^{2}$ $\displaystyle=(a+b)(a+b)^{2}$ $\displaystyle=(a+b)(a^{2}+2ab+b^{2})$ $\displaystyle=(a+b)a^{2}+(a+b)2ab+(a+b)b^{2}$ $\displaystyle=a^{3}+a^{2}b+2a^{2}b+2ab^{2}+ab^{2}+b^{3}$ $\displaystyle=a^{3}+3a^{2}b+3ab^{2}+b^{3}$ (3.57)

There’s a very nice pattern which emerges as we get to higher powers. Let’s try to find a nice expression for how to expand any binomial.

We can use the cases of the expansions of $(a+b)^{1}$, $(a+b)^{2}$ and $(a+b)^{3}$ to guess what form all binomials take. In general, the powers of $a$ seem to start at $a^{n}$ and then go down by one each term. The powers of $b$ seem to do the opposite; they start at $b^{0}$ and increase up to $b^{n}$. In general, we have something which looks a bit like

$(a+b)^{n}=c^{n}_{1}a^{n}+c^{n}_{2}a^{n-1}b+c^{n}_{3}a^{n-2}b^{2}+...+c^{n}_{n-% 1}ab^{n-1}+c^{n}_{n}b^{n}$ (3.58)

Where $c^{n}_{k}$ means the coefficient 66 6 That is to say, the number by which we multiply the variable. e.g. the coefficient of $x^{3}$ in $555x^{3}+14x+15$ would be $555$. of the $k$th item in the expansion of the $n$th power. What are the coefficients, however, and how do we compute them?

Let’s think about what happens to the coefficients when we go from an expansion (e.g. $(a+b)^{3}$ whose value we do know) to an expansion which is one power higher, that we don’t know. We can write this in the "general case" by letting $(a+b)^{n}$ stand for the expansion we do know, and $(a+b)^{n+1}$ for the one we don’t. 77 7 Protip: do this expansion by hand.

The new coefficients clearly depend on the coefficients of the previous binomial 88 8 i.e. for $(a+b)^{n}$ this would be $(a+b)^{n-1}$. But we already know that we can work out the coefficients of the expansion of the binomial of degree $n+1$, if we multiply it by the binomial expansion of $n$. To avoid having to do all that work, it would be better if we could write down the coefficients of all the terms given only some number $n$ without having to multiply out all the brackets by hand.

Using $C^{n}_{k}$ to stand for the $k$th coefficient of the $n$th power, we can also write $(a+b)^{n+1}$ as follows

$(a+b)^{n+1}=C^{n+1}_{0}a^{n+1}+C^{n+1}_{1}a^{n-1}b+...+C^{n+1}_{n}ab^{n}+C^{n+% 1}_{n+1}b^{n+1}$ (3.60)

If we compare $(a+b)^{n+1}$ to $(a+b)^{n}$, the first thing that’s clear is that there’s an $a^{n+1}$ and $b^{n+1}$ which don’t exist in $(a+b)^{n}$. Other than those terms, all the terms in the expansion exist in both $(a+b)^{n}$ and $(a+b)^{n+1}$. Looking at Equation 3.59, we can see that the $k$th term of the expansion of $(a+b)^{n+1}$ (apart from the first and last ones) is equal to

$(c^{n}_{k-1}+c^{n}_{k})$ (3.61)

For example, the coefficient for $a^{n-1}b^{2}$ is $c^{n}_{2}+c^{n}_{3}$ (and as that is the third term in the series, that’s what would be expected.)

In general we can write that

$C^{n+1}_{k}=C^{n}_{k-1}+C^{n}_{k}$ (3.62)

which doesn’t seem very helpful in computing the coefficients.

At this stage, it’s probably easiest to proceed with "proof by divine inspiration" 99 9 This is code for ”the answer is hard to work out, but easy to check”. How this can be proved is explored in the ”Discrete Mathematics” section..

$C^{n}_{k}=\frac{n!}{k!(n-k)!}$ (3.63)

Overall,

$(a+b)^{n}=\sum_{k=0}^{n}C^{n}_{k}a^{n-k}b^{k}$ (3.64)

Note that $C^{n}_{k}$ is also often written as $\binom{n}{k}$ (pronounced "n" choose "k").

###### Example 3.5.1

Find the value of $(x-\sqrt{3})^{4}+(x+\sqrt{3})^{4}$.

These expressions are pretty symmetric (i.e. lots of stuff will cancel when they’re expanded). Expanding the first one gives

 $\displaystyle(x-\sqrt{3})^{4}$ $\displaystyle=x^{4}+4x^{3}(-\sqrt{3})^{1}+6x^{2}(-\sqrt{3})^{2}+4x(-\sqrt{3})^% {3}+(-\sqrt{3})^{4}$ $\displaystyle=x^{4}-4x^{3}\sqrt{3}+6\cdot 3x^{2}-4x\sqrt{3}^{3}+3^{2}$ $\displaystyle=x^{4}-4x^{3}\sqrt{3}+18x^{2}-4x\sqrt{3}^{3}+9$

Note that because the other expression contains no negative numbers, everything that was negative in the expansion of $(x-\sqrt{3})^{4}$ will instead be positive, so

$(x+\sqrt{3})^{4}=x^{4}+4x^{3}\sqrt{3}+18x^{2}+4x\sqrt{3}^{3}+9$

and thus that overall,

$(x-\sqrt{3})^{4}+(x+\sqrt{3})^{4}=2x^{4}+36x^{2}+18$
###### Example 3.5.2

Find the sum of all the coefficients in the expression $(1+x)^{n}$.

Expanding this,

$(1+x)^{n}=1+\binom{n}{1}x+\binom{n}{2}x^{2}+...+\binom{n}{n}x^{n}$

Then, set $x=1$ in this expression 1010 10 This ”trick” is something that also shows up in other places. e.g. partial fractions (see the partial fractions section of the ”algebra” topic for details)..

 $\displaystyle 2^{n}=1+\binom{n}{1}+\binom{n}{2}+...+\binom{n}{n}$ $\displaystyle 2^{n}-1=\binom{n}{1}+\binom{n}{2}+...+\binom{n}{n}$ (3.65)