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


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

(a+b)2\displaystyle(a+b)^{2} =(a+b)(a+b)\displaystyle=(a+b)(a+b)
=a2+2ab+b2\displaystyle=a^{2}+2ab+b^{2} (3.56)

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

(a+b)2\displaystyle(a+b)^{2} =(a+b)(a+b)2\displaystyle=(a+b)(a+b)^{2}
=a3+3a2b+3ab2+b3\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)^{1}, (a+b)2(a+b)^{2} and (a+b)3(a+b)^{3} to guess what form all binomials take. In general, the powers of aa seem to start at ana^{n} and then go down by one each term. The powers of bb seem to do the opposite; they start at b0b^{0} and increase up to bnb^{n}. In general, we have something which looks a bit like

(a+b)n=c1nan+c2nan1b+c3nan2b2++cn1nabn1+cnnbn(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 cknc^{n}_{k} means the coefficient 66 6 That is to say, the number by which we multiply the variable. e.g. the coefficient of x3x^{3} in 555x3+14x+15555x^{3}+14x+15 would be 555555. of the kkth item in the expansion of the nnth 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(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(a+b)^{n} stand for the expansion we do know, and (a+b)n+1(a+b)^{n+1} for the one we don’t. 77 7 Protip: do this expansion by hand.

(a+b)n+1\displaystyle(a+b)^{n+1} =(a+b)(a+b)n\displaystyle=(a+b)(a+b)^{n}
=(a+b)[c1nan+c2nan1b+c3nan2b2++cn2na2bn2\displaystyle=(a+b)[c^{n}_{1}a^{n}+c^{n}_{2}a^{n-1}b+c^{n}_{3}a^{n-2}b^{2}+...% +c^{n}_{n-2}a^{2}b{n-2}
=[c1nan+1+c2nanb+c3nan1b2++cn2na3bn2\displaystyle=[c^{n}_{1}a^{n+1}+c^{n}_{2}a^{n}b+c^{n}_{3}a^{n-1}b^{2}+...+c^{n% }_{n-2}a^{3}b{n-2}
+[c1nanb+c2nan1b2+c3nan2b3++cn2na2bn1+cn1nabn\displaystyle\quad+[c^{n}_{1}a^{n}b+c^{n}_{2}a^{n-1}b^{2}+c^{n}_{3}a^{n-2}b^{3% }+...+c^{n}_{n-2}a^{2}b{n-1}+c^{n}_{n-1}ab^{n}
=c1nan+1+(c1n+c2n)anb+(c2n+c3n)an1b2+\displaystyle=c^{n}_{1}a^{n+1}+(c^{n}_{1}+c^{n}_{2})a^{n}b+(c^{n}_{2}+c^{n}_{3% })a^{n-1}b^{2}+...
+(cn2n+cn1n)a2bn1+(cn1n+cnn)abn+cnnbn+1\displaystyle\quad+(c^{n}_{n-2}+c^{n}_{n-1})a^{2}b^{n-1}+(c^{n}_{n-1}+c^{n}_{n% })ab^{n}+c^{n}_{n}b^{n+1} (3.59)

The new coefficients clearly depend on the coefficients of the previous binomial 88 8 i.e. for (a+b)n(a+b)^{n} this would be (a+b)n1(a+b)^{n-1}. But we already know that we can work out the coefficients of the expansion of the binomial of degree n+1n+1, if we multiply it by the binomial expansion of nn. 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 nn without having to multiply out all the brackets by hand.

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

(a+b)n+1=C0n+1an+1+C1n+1an1b++Cnn+1abn+Cn+1n+1bn+1(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(a+b)^{n+1} to (a+b)n(a+b)^{n}, the first thing that’s clear is that there’s an an+1a^{n+1} and bn+1b^{n+1} which don’t exist in (a+b)n(a+b)^{n}. Other than those terms, all the terms in the expansion exist in both (a+b)n(a+b)^{n} and (a+b)n+1(a+b)^{n+1}. Looking at Equation 3.59, we can see that the kkth term of the expansion of (a+b)n+1(a+b)^{n+1} (apart from the first and last ones) is equal to

(ck1n+ckn)(c^{n}_{k-1}+c^{n}_{k}) (3.61)

For example, the coefficient for an1b2a^{n-1}b^{2} is c2n+c3nc^{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

Ckn+1=Ck1n+CknC^{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..

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


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

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

Example 3.5.1

Find the value of (x3)4+(x+3)4(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

(x3)4\displaystyle(x-\sqrt{3})^{4} =x4+4x3(3)1+6x2(3)2+4x(3)3+(3)4\displaystyle=x^{4}+4x^{3}(-\sqrt{3})^{1}+6x^{2}(-\sqrt{3})^{2}+4x(-\sqrt{3})^% {3}+(-\sqrt{3})^{4}
=x44x33+63x24x33+32\displaystyle=x^{4}-4x^{3}\sqrt{3}+6\cdot 3x^{2}-4x\sqrt{3}^{3}+3^{2}

Note that because the other expression contains no negative numbers, everything that was negative in the expansion of (x3)4(x-\sqrt{3})^{4} will instead be positive, so


and thus that overall,

Example 3.5.2

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

Expanding this,


Then, set x=1x=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)..

2n=1+(n1)+(n2)++(nn)\displaystyle 2^{n}=1+\binom{n}{1}+\binom{n}{2}+...+\binom{n}{n}
2n1=(n1)+(n2)++(nn)\displaystyle 2^{n}-1=\binom{n}{1}+\binom{n}{2}+...+\binom{n}{n} (3.65)