The Binomial Theorem

Starting point: what you already know

• Expanding brackets: $(a+b{)}^2=a^2+2ab+b^2$.
• Index laws: $(2x{)}^3=8x^3$ and $x^a\times x^b=x^{a+b}$.
• Combinations: $\left(\genfrac{}{}{0ex}{}{n}{r}\right)=\frac{n!}{r!(n-r)!}$ counts ways to choose $r$ things from $n$ (previous chapter).

Pascal's triangle

Expand a few powers and look at the numbers in front:

Each row starts and ends with 1, and each inside number is the sum of the two numbers above it. This is Pascal's triangle. The next row is 1 5 10 10 5 1.

Why combinations appear

$(a+b{)}^4=(a+b)(a+b)(a+b)(a+b)$. To get an $a^2{b}^2$ term, you pick $b$ from 2 of the 4 brackets and $a$ from the rest. There are $\left(\genfrac{}{}{0ex}{}{4}{2}\right)=6$ ways to choose those brackets, so the coefficient is 6.

Patterns in every expansion:

• There are $n+1$ terms.
• The power of $a$ goes down from $n$ to 0 while the power of $b$ goes up from 0 to $n$.
• The powers in each term add to $n$.

The general term

The term with $b^r$ is

$$T_{r+1}=\left(\genfrac{}{}{0ex}{}{n}{r}\right)a^{n-r}{b}^r$$

It is called $T_{r+1}$ because the first term has $r=0$.

Quick summary

• Pascal's triangle: each number is the sum of the two above. Row $n$ gives the coefficients of $(a+b{)}^n$.
• $(a+b{)}^n=\sum _{r=0}^n(\genfrac{}{}{0ex}{}{n}{r})a^{n-r}{b}^r$, with $n+1$ terms.
• General term $T_{r+1}=\left(\genfrac{}{}{0ex}{}{n}{r}\right)a^{n-r}{b}^r$. Set the power of $x$ to what you need and solve for $r$.
• The sum of row $n$ of Pascal's triangle is $2^n$.

Quiz