# A calculus of the absurd

#### 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. $$(a+b)^2$$ 66 If you’ve no idea how to expand this, review a GCSE textbook. can be expanded like this

\begin{align} (a+b)^2 &= (a+b)(a+b) \notag \\ &= (a+b)a + (a+b)b \notag \\ &= a^2 + ab + ab + b^2 \notag \\ &= a^2 + 2ab + b^2 \end{align}

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

\begin{align} (a+b)^2 &= (a+b)(a+b)^2 \notag \\ &= (a+b)(a^2 + 2ab + b^2) \notag \\ &= (a+b)a^2 + (a+b)2ab + (a+b)b^2 \notag \\ &= a^3 + a^2b + 2a^2b + 2ab^2 + ab^2 + b^3 \notag \\ &= a^3 + 3a^2b + 3ab^2 + b^3 \end{align}

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

$$(a+b)^n = c^{n}_1 a^n + c^{n}_2 a^{n-1}b + c^{n}_3 a^{n-2}b^2 + ... + c^{n}_{n-1} ab^{n-1} + c^{n}_n b^n$$

Where $$c^{n}_k$$ means the coefficient 77 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. 88 Protip: do this expansion by hand.

The new coefficients clearly depend on the coefficients of the previous binomial 99 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

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

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 3.43, 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

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

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

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

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

At this stage, it’s probably easiest to proceed with "proof by divine inspiration" 1010 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..

$$\label {binomial2factorial} C^{n}_{k} = \frac {n!}{k!(n-k)!}$$

Overall,

$$\label {binomial expansion} (a + b)^n = \sum _{k=0}^{n} C^{n}_{k} a^{n-k} b^k$$

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

• Example 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

\begin{align*} (x-\sqrt {3})^4 &= x^4 + 4 x^3 (-\sqrt {3})^1 + 6 x^2 (-\sqrt {3})^2 + 4 x (-\sqrt {3})^3 + (-\sqrt {3})^4 \\ &= x^4 - 4 x^3 \sqrt {3} + 6 \cdot 3 x^2 - 4 x \sqrt {3}^3 + 3^2 \\ &= x^4 - 4 x^3 \sqrt {3} + 18 x^2 - 4 x \sqrt {3}^3 + 9 \end{align*}

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

\begin{equation*} (x+\sqrt {3})^4 = x^4 + 4 x^3 \sqrt {3} + 18 x^2 + 4 x \sqrt {3}^3 + 9 \end{equation*}

and thus that overall,

\begin{equation*} (x - \sqrt {3})^4 + (x + \sqrt {3})^4 = 2 x^4 + 36 x^2 + 18 \end{equation*}

• Example 2 Find the sum of all the coefficients in the expression $$(1+x)^n$$.

Expanding this,

\begin{equation*} (1 + x)^n = 1 + \binom {n}{1} x + \binom {n}{2} x^2 + ... + \binom {n}{n} x^n \end{equation*}

Then, set $$x=1$$ in this expression 1111 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)..

\begin{align} & 2^n = 1 + \binom {n}{1} + \binom {n}{2} + ... + \binom {n}{n} \notag \\ & 2^n - 1 = \binom {n}{1} + \binom {n}{2} + ... + \binom {n}{n} \end{align}