At O-Level you expand (a+b)2 and (a+b)3 by multiplying out brackets. For a power such as (a+b)12 that is far too slow — the binomial theorem writes down every term of the expansion directly. For a positive integer n, (a+b)n=an+ n1an-1b+ n2an-2b2+. s+ nran-rbr+. s+bn. **Features of the expansion of (a+b)n** - There are n+1 terms. - The powers of a descend from n to 0. - The powers of b ascend from 0 to n. - In every term, the powers of a and b add up to n. **Evaluating nr** **Notation** | **Definition** | **Example** Factorial | n!=n x (n-1) x. s x 2 x 1 | 6!=6 x 5 x 4 x 3 x 2 x 1 =720 Combination | nr= n!r!,(n-r)!= n(n-1). s(n-r+1)r! | 82= 8 x 72!=28 Calculator | nr=nC_r | 103=10C_3=120 **Special values of nr** **Result** | **Example** 0!=1 | nn= n!n!,0!=1 n0=1 | 70=1 n1=n | 71=7 nr= nn-r | 75= 72=21 r is an integer with 0 <= r <= n. If your equation for r gives a value outside this range, or a non-integer, the required term does not exist in the expansion.

The Binomial Theorem

Secondary 4

TGM Original Questions

From O-Level to Additional Mathematics

At O-Level you expand \((a+b)^2\) and \((a+b)^3\) by multiplying out brackets. For a power such as \((a+b)^{12}\) that is far too slow — the binomial theorem writes down every term of the expansion directly.

The binomial theorem

For a positive integer \(n\), \((a+b)^n=a^n+\dbinom{n}{1}a^{n-1}b+\dbinom{n}{2}a^{n-2}b^2+\cdots+\dbinom{n}{r}a^{n-r}b^r+\cdots+b^n\).

Features of the expansion of \((a+b)^n\)

There are \(n+1\) terms.
The powers of \(a\) descend from \(n\) to \(0\).
The powers of \(b\) ascend from \(0\) to \(n\).
In every term, the powers of \(a\) and \(b\) add up to \(n\).

Evaluating \(\dbinom{n}{r}\)

Notation	Definition	Example
Factorial	\(n!=n\times(n-1)\times\cdots\times 2\times 1\)	\(6!=6\times5\times4\times3\times2\times1\) \(=720\)
Combination	\(\dbinom{n}{r}=\dfrac{n!}{r!\,(n-r)!}=\dfrac{n(n-1)\cdots(n-r+1)}{r!}\)	\(\dbinom{8}{2}=\dfrac{8\times7}{2!}=28\)
Calculator	\(\dbinom{n}{r}={}^{n}C_{r}\)	\(\dbinom{10}{3}={}^{10}C_{3}=120\)

Special values of \(\dbinom{n}{r}\)

Result	Example
\(0!=1\)	\(\dbinom{n}{n}=\dfrac{n!}{n!\,0!}=1\)
\(\dbinom{n}{0}=1\)	\(\dbinom{7}{0}=1\)
\(\dbinom{n}{1}=n\)	\(\dbinom{7}{1}=7\)
\(\dbinom{n}{r}=\dbinom{n}{n-r}\)	\(\dbinom{7}{5}=\dbinom{7}{2}=21\)

Exam check

\(r\) is an integer with \(0\le r\le n\). If your equation for \(r\) gives a value outside this range, or a non-integer, the required term does not exist in the expansion.

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