Maths › Sequences and series › The general binomial expansion
The general binomial expansion
Let the exponent be negative or fractional and the binomial theorem keeps going, but the expansion never stops. What you get is an infinite series that is valid only inside a window of x-values. Learn where the window is and the series will compute square roots and reciprocals to whatever accuracy you want.
Builds on The binomial expansion and Partial fractions.
IN THIS TOPIC
- Expand (1 + x)n for negative and fractional n, as far as a stated power.
- State the range of validity, and say why the series is silent outside it.
- Handle (a + bx)n by taking an out of the bracket first.
- Combine expansions with partial fractions, and choose the right window for the sum.
COMMON MISCONCEPTION
The expansion of (1 + x)−1 works for every x.
The series that never ends
For a positive whole n the expansion stops, because one coefficient eventually reaches zero and every term after it is zero too. Make n negative or fractional and nothing ever cancels. The same booklet formula then runs forever,
and it converges throughout the window |x| < 1. Beyond it, with |x| > 1, the terms eventually grow instead of shrinking and the series diverges; at the endpoints x = ±1 themselves the behaviour depends on n and has to be checked separately. The window is not decoration. Feed x = 1 into the expansion of (1 + x)−1 and the running totals go 1, 0, 1, 0, 1, never settling anywhere near the true value of ½.
WORKED EXAMPLE
A square root, expanded
Expand (1 − 2x)1/2 in ascending powers of x up to the x2 term, and state the range of validity.
Use the formula with n = ½ and x replaced by −2x throughout. That gives 1 + ½(−2x) + [½ × (−½)/2](−2x)2.
The terms tidy to 1 − x − x2/2.
Validity needs |−2x| < 1, so |x| < ½.
The replacement goes in whole, sign and coefficient together, and the window shrinks to match. Substituting only the x and leaving the −2 behind is the error that wrecks these questions from line one.
Brackets that do not start at 1
The formula needs a 1 at the front, so produce one. Pull an outside and expand what is left over,
which is valid for |bx/a| < 1. The constant is raised to the power n on its way out, and then it multiplies every single term of the expansion that follows.
GUIDED PRACTICE
Factor out, then expand
Expand (4 + x)1/2 up to the x2 term, stating the validity, before opening the working.
Show the working
Factor first. (4 + x)1/2 = 41/2(1 + x/4)1/2 = 2(1 + x/4)1/2.
Expand: 2[1 + ½(x/4) + (½ × (−½)/2)(x/4)2] = 2 + x/4 − x2/64.
Validity: |x/4| < 1, so |x| < 4.
Notice the 4 left as 41/2 = 2, not as 4. Forgetting to raise the factored constant to the power is the single biggest source of dropped marks in this topic.
Partial fractions, and whose window wins
Partial fractions are what make this possible. Split an ugly rational function into linear brackets, expand each bracket by this lesson's formula, and add the results term by term. Every piece arrives with its own validity window. The combined expansion is valid only where all of them hold at once, so take the tightest.
INDEPENDENT PRACTICE
Partial fractions feed the series
Using the decomposition (5x + 7)/((x + 1)(x + 2)) = 2/(x + 1) + 3/(x + 2), expand the original fraction up to the x2 term and state the validity.
Show the working
Each piece expands separately. 2(1 + x)−1 = 2 − 2x + 2x2 − …, and 3/(x + 2) = (3/2)(1 + x/2)−1 = 3/2 − 3x/4 + 3x2/8 − ….
Adding gives 7/2 − 11x/4 + 19x2/8.
The windows are |x| < 1 and |x| < 2. Only the tighter one survives, so the expansion is valid for |x| < 1.
A decimal check at x = 0.05 puts both sides at 3.368 to 4 significant figures. One divergent piece poisons the whole sum, so the strictest window always wins.
ASSESSMENT FOCUS
- Quote the general formula before substituting, and put the whole of bx in, sign and coefficient together.
- Stop at the power the question names. Extra terms are not asked for and cost minutes.
- Take an out of (a + bx)n first, and remember it multiplies every term afterwards.
- The validity statement is a mark in its own right. Solve |bx/a| < 1 and write the answer as an inequality in x.
- For a combined expansion, report the tightest window of the pieces, and say that is what you have done.
CHECK YOURSELF
Expand (1 + 3x)−2 in ascending powers up to x2, state the validity, and use the expansion to estimate 1.03−2.
Show a hint
n = −2, and the replacement is 3x; then x = 0.01.
Show the answer
1 + (−2)(3x) + [(−2)(−3)/2](3x)2 = 1 − 6x + 27x2, valid for |x| < ⅓.
At x = 0.01 the truncation gives 1 − 0.06 + 0.0027 = 0.9427.
The true value is 0.94260…, so three significant figures agree and the fourth is out by one. The discarded cubic term is the whole of that gap.
A negative or non-integer exponent makes the expansion infinite; a positive whole number is rational too, and terminates. |bx/a| < 1 is where the infinite one converges, with the endpoints needing their own check.
Take out the a, expand the rest, and never leave the validity unsaid.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
Or read them with their worked answers on the the general binomial expansion questions page.
CHECK YOUR PROGRESS
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- Expand (1 + x)n for negative and fractional n, as far as a stated power.
- State the range of validity, and say why the series is silent outside it.
- Handle (a + bx)n by taking an out of the bracket first.
- Combine expansions with partial fractions, and choose the right window for the sum.
Open the full revision checklist to see every objective in the course in one place.