The Binomial Series

Chapter 6 student notes

1. The binomial series

Write enough terms symbolically before simplifying; that prevents sign and denominator errors.

y = (1+x)⁻¹ᐟ²third-order seriesThe approximation is centred at x = 0 and valid for |x| < 1.
General form
\[(1+x)^n=1+nx+\frac{n(n-1)}{2!}x^2+\frac{n(n-1)(n-2)}{3!}x^3+\cdots\]
General coefficient
\[\binom nr=\frac{n(n-1)\cdots(n-r+1)}{r!}\]
Validity
\[n otin\mathbb Z_{\ge0}\Rightarrow |x|<1\]

Rational index expansion

Expand \((1-2x)^{-1/2}\) to \(x^3\): \[1+\left(-\frac12\right)(-2x)+\frac{(-1/2)(-3/2)}2(-2x)^2+\frac{(-1/2)(-3/2)(-5/2)}6(-2x)^3\] \[=1+x+\frac32x^2+\frac52x^3+\cdots.\]
Common mistake. The sign and power belong to the entire \(x\)-term. For example \((-x/4)^2=+x^2/16\).

2. Rewriting before expansion

Most nonstandard binomial questions are standard after one careful factorisation.

  1. Factor out a constant so the bracket becomes \(1+u\).
  2. Identify \(n\) and \(u\) exactly.
  3. Expand to the order needed.
  4. Multiply back any outside factor.
  5. State the validity range obtained from \(|u|<1\) when required.

Rewrite pattern

\[(9-2x)^{-1/2}=9^{-1/2}\left(1-\frac{2x}{9}\right)^{-1/2}=\frac13\left(1-\frac{2x}{9}\right)^{-1/2}.\] Now use \(u=-2x/9\); the series is valid when \(|2x/9|<1\), i.e. \(|x|<9/2\).

3. Coefficient and parameter questions

Coefficient questions reward controlled truncation rather than brute-force expansion.

Coefficient extraction

In \((1+2x)^8\), the coefficient of \(x^3\) is \[\binom83 2^3=56\cdot8=448.\]

Product of expansions

If you need the coefficient of \(x^2\) in \((1+x)^p(1-2x)^q\), expand each factor only to \(x^2\), then combine contributions from constant×\(x^2\), \(x\)×\(x\), and \(x^2\)×constant.
Exam tip. Do not expand further than the requested power. “Up to and including \(x^3\)” means terms through degree 3 only.

4. Surd and numerical approximation

Binomial approximation links exact series algebra to fast numerical estimates.

How the approximation works

To approximate \(\sqrt{1.02}\), write \[\sqrt{1.02}=(1+0.02)^{1/2}\] and substitute \(x=0.02\) into a short binomial expansion. The small size of \(|x|\) makes the neglected higher powers small.
Exam tip. Use the exact requested substitution and keep enough terms to justify the required decimal accuracy. A “hence” part usually expects reuse of the expansion you just found.

5. Chapter mastery checklist