Expand any power of a bracket, or pick out just one term.

Pascal's triangle and the full expansion.

Expand \((a + b)^{2} = a^{2} + 2ab + b^{2}\) and \((a + b)^{3} = a^{3} + 3a^{2}b + 3ab^{2} + b^{3}\) and the coefficients 1, 2, 1 and 1, 3, 3, 1 are rows of Pascal's triangle (Fig. 1). This is no accident: the coefficient of \(a^{n-r}b^{r}\) counts the ways of choosing \(b\) from \(r\) of the \(n\) brackets, which is \({}^{n}C_{r}\). That gives the binomial theorem, printed in the List of formulas:

\[ (a + b)^{n} = a^{n} + {}^{n}C_{1}\,a^{n-1}b + {}^{n}C_{2}\,a^{n-2}b^{2} + \cdots + b^{n} \]

• Number of terms: \(n + 1\), from \(r = 0\) to \(r = n\).
• Powers: the power of \(a\) falls from \(n\) to 0 while the power of \(b\) rises from 0 to \(n\); in every term they add up to \(n\).
• Coefficients: row \(n\) of Pascal's triangle for small \(n\), or \({}^{n}C_{r}\) worked out by cancelling for larger \(n\).
• Substitute whole terms: for \((2 - 3x)^{5}\), \(a = 2\) and \(b = (-3x)\), sign and number included, in brackets.
• Ascending powers of \(x\): start with the constant term, then the \(x\) term, then \(x^{2}\), and so on.

The general term: find one coefficient.

When a question asks for a single term, such as the coefficient of \(x^{4}\) in \(\left(3 - \dfrac{x}{2}\right)^{8}\), you do not need the whole expansion. Every term has the same shape, so write down the general term, find the power of \(x\) in it, and choose \(r\) to give the power you want.

\[ T_{r+1} = {}^{n}C_{r}\, a^{n-r} b^{r}, \qquad 0 \le r \le n \]

• Which term: \(r = 0\) gives the 1st term, so \({}^{n}C_{r}\,a^{n-r}b^{r}\) is the \((r + 1)\)th term.
• Power of \(x\): collect the powers of \(x\) from \(a^{n-r}\) and from \(b^{r}\) into one expression in \(r\).
• Independent of \(x\): the constant term, where that power of \(x\) is 0. With \(x\) and \(\dfrac{1}{x}\) or \(\dfrac{1}{x^{2}}\) in the bracket, the negative powers cancel some of the positive ones.
• Check: \(r\) must be a whole number from 0 to \(n\); if it is not, there is no such term.

Unknown constants, products and approximations.

Exam questions often run the general term backwards: you are told a coefficient and must find a constant in the bracket, or the power \(n\) itself. Others multiply an expansion by a second bracket, so a term in the answer can be built in more than one way.

• Given a coefficient: write the coefficient in terms of the unknown, set it equal to the value and solve. A square such as \(k^{2} = 4\) gives \(k = \pm 2\) unless the question says \(k\) is positive.
• Given two coefficients: form two equations; dividing one by the other often removes an unknown.
• Products: for the \(x^{2}\) term of \((p + qx)(\dots)\), pair \(p\) with the \(x^{2}\) term and \(qx\) with the \(x\) term of the expansion, then add.
• Approximations: choose \(x\) so that the bracket matches the number, such as \((1 + x)^{5}\) with \(x = 0.02\) for \(1.02^{5}\); the first few terms give a close value when \(x\) is small.