Topic 12 Binomial Theorem
12.1 Pascal’s triangle
Binomial Theorem gives a formula for \((a+b)^n\). First few:
- \((a+b)^2 = a^2+2ab+b^2\)
- \((a+b)^3 = (a^2+2ab+b^2)(a+b) = a^3+3a^2b+3ab^2+b^3\)
- \((a+b)^4 = \dots = a^4+4a^3b+6a^2b^2 + 4ab^3 + b^4\)
In general: \[ (a+b)^n = C_{n,0}\cdot a^n + C_{n,1}\cdot a^{n-1}b + C_{n,2}\cdot a^{n-2}b^2 + \dots + C_{n,n-1}\cdot ab^{n-1} + C_{n,n}\cdot b^n \] for some coefficients \(C_{n,0},C_{n,1},\dots,C_{n,n}\). “Extrapolating” from the examples above, we can “guess”: \[\begin{align*} C_{n,0} &= C_{n,n} = 1\\ C_{n,1} &= C_{n,n-1} = n\\ C_{n,k} &= C_{n-1,k-1}+C_{n-1,k} \end{align*}\]
The two rules: \(C_{n,0}=C_{n,n}=1\) and \(C_{n,k}=C_{n-1,k-1}+C_{n-1,k}\) are the building laws for Pascal’s triangle: \[ \begin{array}{cccccccccccccc} & & & & & & &1 & & & & & & \\ & & & & & &1 & &1 & & & & & \\ & & & & &1 & &2 & &1 & & & & \\ & & & &1 & &3 & &3 & &1 & & & \\ & & &1 & &4 & &6 & &4 & &1 & & \\ & &1 & &5 & &10& &10& &5 & &1 & \\ &1 & &6 & &15& &20& &15& &6 & &1 \\ & & & & & & &\vdots& & & & & & \\ \end{array} \] ::: {.example} \((a+b)^6 = a^6 + 6 a^5b + 15 a^4b^2 + 20 a^3b^3 + 15 a^2b^4 + 6 ab^5 + b^6\). ::: ::: {.example} \[\begin{align*}(a+2x)^4 &= a^4 + 4a^3(2x) + 6a^2(2x)^2 + 4a(2x)^3+(2x)^4\\ &= a^4 +8a^3x+24a^2x^2+32ax^3+16x^4. \end{align*}\] :::
Example 12.1 Expand \((1.005)^4\) to four decimal places (4 d.p).
Solution. \[\begin{align*} (1.005)^4 &= (1+0.005)^4\\ &= 1+ 4\cdot 0.005 + 6\cdot (0.005)^2 + \underbrace{4\cdot(0.005)^3 + (0.005)^4}_{\text{do not contribute to 4 d.p.}}\\ &= 1+0.02 + 0.00015 + \dots \\ &\approx 1.0202\quad\quad\text{(to 4 d.p.)} \end{align*}\]
12.2 Binomial coefficients
For a positive integer \(m\), define \(m\) factorial, denoted “\(m!\)”, as
\(m!=1\cdot2\cdot 3\cdots (m-1)\cdot m\);
and declare that \(0!=1\).
Theorem 12.1 \(\boxed{C_{n,k} = \dfrac{n!}{k!\cdot (n-k)!}}\)
Note: The number \(C_{n,k}\) is also denoted by \(\displaystyle{n\choose k}\), read ``\(n\) choose \(k\)’’2.
Proof (non–examinable): To argue that the formula “works correctly”, it suffices to check that the number above satisfies the laws defining Pascal’s triangle.
That \(C_{n,0}=C_{n,n}=1\) is clear.
Now checking that \(C_{n,k}=C_{n-1,k-1}+C_{n-1,k}\): \[ \begin{matrix} \dfrac{n!}{k!(n-k)!} &\stackrel{?}{=} & \dfrac{(n-1)!}{(k-1)!(n-k)!} & + & \dfrac{(n-1)!}{k!(n-1-k)!}\\ & & = & & = \\ & & \dfrac{k}{n}\cdot\dfrac{n!}{k!(n-k)!} & + & \dfrac{n-k}{n}\cdot \dfrac{n!}{k!(n-k)!} \end{matrix} \] This finishes the proof.
Let us analyse the formula: observe that \(C_{n,k} = \dfrac{n(n-1)(n-2)\cdots(n-k+1)}{1\cdot 2\cdot 3 \cdots k}\).
We will use this expression as a definition for \(C_{n,k}\) when \(n\) is negative or a fraction.
Theorem 12.2 If \(-1<x<1\), then the right hand side of \[\begin{align*} (1+x)^n &= C_{n,0}+C_{n,1}x+C_{n,2}x^2 + C_{n,3}x^3 + \dots\\ &= 1+ nx + \frac{n(n-1)}{2}x^2 + \frac{n(n-1)(n-2)}{6} x^3 + \dots \end{align*}\] converges3 and this equality is valid also when \(n\) is negative or a fraction.
Example 12.2 \(\dfrac{1}{1+x} = (1+x)^{-1}\) \(=1-x+x^2-x^3+x^4 \mp\cdots\)
Example 12.3 \(\dfrac{1}{1-x} = (1-x)^{-1} = 1+x+x^2+x^3+\dots\)
Example 12.4 Expand \(\dfrac{1}{(1+x)^3}\) up to the term in \(x^4\).
Solution. \[\begin{align*} \frac{1}{(1+x)^3} &= (1+x)^{-3}\\ &= 1+(-3)x+\frac{(-3)(-4)}{1\cdot 2}x^2 + \frac{(-3)(-4)(-5)}{1\cdot 2\cdot 3}x^3 + \frac{(-3)(-4)(-5)(-6)}{1\cdot 2\cdot 3\cdot 4}x^4 + \cdots\\ &=1-3x+6x^2-10x^3+15x^4 \mp \cdots \end{align*}\]
Example 12.5 Expand \((3-x)^{-4}\) up to the term in \(x^3\).
Solution. \[\begin{align*} (3-x)^{-4} &= 3^{-4}\left(1-\frac{x}{3}\right)^{-4}\\ &= 3^{-4}\left(1+(-4)\left(-\frac{x}{3}\right) +\frac{(-4)(-5)}{1\cdot 2}\left(-\frac{x}{3}\right)^2 + \frac{(-4)(-5)(-6)}{1\cdot 2\cdot 3}\left(-\frac{x}{3}\right)^3+\cdots\right)\\ &= \frac{1}{81}\left(1+\frac{4}{3}x+\frac{10}{9}x^2+\frac{20}{27}x^3+\cdots\right) \end{align*}\] converges and equality is valid for \(-1<\frac{x}{3}<1\) \(\iff\) \(-3<x<3\).
This is because it also computes the number of ways one can choose \(k\) objects out of \(n\).↩︎
“Converges” is a mathematical term that you will learn more precisely later on; the meaning here is roughly that “for each \(x\), the infinite sum does describe a well-defined and unique number, and we get better and better approximations to it by adding up more and more terms of the infinite series”.↩︎