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How do we sum arithmetic and geometric series, handle infinite sums, and expand a binomial for any rational power?

Arithmetic and geometric sequences and series, sigma notation, the sum to infinity of a convergent geometric series, and the binomial expansion for any rational index.

A focused answer to WJEC A2 Unit 3 sequences and series, covering arithmetic and geometric sequences and series, sigma notation, the sum to infinity of a convergent geometric series, and the binomial expansion for any rational index.

Generated by Claude Opus 4.813 min answer

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  1. What this dot point is asking
  2. The answer
  3. Examples in context
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What this dot point is asking

WJEC wants you to work with arithmetic and geometric sequences and series, to use sigma notation, to find the sum to infinity of a convergent geometric series, and to expand (1+x)n(1 + x)^n as a binomial series for any rational index nn (with its validity condition). The general binomial series extends the AS work, where nn had to be a positive integer.

The answer

Arithmetic sequences and series

An arithmetic sequence has a constant common difference dd. The nnth term is a+(n1)da + (n-1)d.

Geometric sequences and series

A geometric sequence has a constant common ratio rr. The nnth term is arn1ar^{n-1}.

Sigma notation

The symbol \sum denotes a sum: r=1nur=u1+u2++un\displaystyle\sum_{r=1}^{n} u_r = u_1 + u_2 + \cdots + u_n. The lower and upper limits give the first and last index, and the expression after \sum is the general term. A useful property is that \sum is linear: (aur+b)=aur+bn\displaystyle\sum (au_r + b) = a\sum u_r + bn, so constant multiples factor out and a constant term added nn times contributes bnbn. For example r=14(2r+1)=2(1+2+3+4)+4(1)=20+4=24\displaystyle\sum_{r=1}^{4} (2r + 1) = 2(1+2+3+4) + 4(1) = 20 + 4 = 24, which you can also check by listing the terms 3,5,7,93, 5, 7, 9.

The binomial series for any index

For any rational nn, the expansion is an infinite series valid for x<1|x| < 1:

Examples in context

Example 1. A recurring decimal. The decimal 0.44440.4444\ldots is the geometric series 0.4+0.04+0.004+0.4 + 0.04 + 0.004 + \cdots with a=0.4a = 0.4 and r=0.1r = 0.1. Its sum to infinity is 0.410.1=0.40.9=49\dfrac{0.4}{1 - 0.1} = \dfrac{0.4}{0.9} = \dfrac{4}{9}, confirming the fraction. The sum to infinity turns a recurring decimal into an exact fraction.

Example 2. Approximating a root. Using (1+x)1/21+12x18x2(1 + x)^{1/2} \approx 1 + \tfrac{1}{2}x - \tfrac{1}{8}x^2 with x=0.04x = 0.04 gives 1.041+0.020.00002=1.0198\sqrt{1.04} \approx 1 + 0.02 - 0.00002 = 1.0198, very close to the true 1.01981.0198. The binomial series gives quick approximations to roots near 11.

Try this

Q1. An arithmetic series has first term 55 and common difference 33. Find the sum of the first 10 terms. [3 marks]

  • Cue. S10=102[2(5)+9(3)]=5(10+27)=185S_{10} = \tfrac{10}{2}[2(5) + 9(3)] = 5(10 + 27) = 185.

Q2. A geometric series has a=100a = 100 and r=0.2r = 0.2. Find the sum to infinity. [2 marks]

  • Cue. r<1|r| < 1, so S=10010.2=1000.8=125S_\infty = \dfrac{100}{1 - 0.2} = \dfrac{100}{0.8} = 125.

Q3. Find the term in x2x^2 in the expansion of (1x)1(1 - x)^{-1}. [2 marks]

  • Cue. (1x)1=1+x+x2+(1 - x)^{-1} = 1 + x + x^2 + \cdots, so the x2x^2 term is x2x^2 (coefficient 11).

Q4. State the validity range for the expansion of (1+3x)1(1 + 3x)^{-1}. [1 mark]

  • Cue. Valid for 3x<1|3x| < 1, that is x<13|x| < \tfrac{1}{3}.

Exam-style practice questions

Practice questions written in the style of WJEC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

WJEC A2 style5 marksA geometric series has first term 88 and common ratio 34\dfrac{3}{4}. Find its sum to infinity, and the sum of the first 5 terms.
Show worked answer →

For a convergent geometric series use the sum-to-infinity formula, and for a finite sum use the partial-sum formula.

Sum to infinity (valid since r=34<1|r| = \tfrac{3}{4} < 1): S=a1r=8134=814=32S_\infty = \dfrac{a}{1 - r} = \dfrac{8}{1 - \tfrac{3}{4}} = \dfrac{8}{\tfrac{1}{4}} = 32.

First 5 terms: S5=a(1r5)1r=8(1(34)5)14S_5 = \dfrac{a(1 - r^5)}{1 - r} = \dfrac{8\left(1 - (\tfrac{3}{4})^5\right)}{\tfrac{1}{4}}.

(34)5=24310240.2373(\tfrac{3}{4})^5 = \dfrac{243}{1024} \approx 0.2373, so S5=32(10.2373)=32×0.762724.4S_5 = 32(1 - 0.2373) = 32 \times 0.7627 \approx 24.4.

Markers reward checking r<1|r| < 1 for the sum to infinity, the correct value 3232, and the partial-sum formula giving about 24.424.4. Using the sum to infinity formula when r1|r| \ge 1 is a serious error.

WJEC A2 style5 marksFind the first three terms in the binomial expansion of (1+2x)2(1 + 2x)^{-2}, and state the values of xx for which it is valid.
Show worked answer →

Use the general binomial series for any index nn, with n=2n = -2.

(1+y)n=1+ny+n(n1)2!y2+(1 + y)^n = 1 + ny + \dfrac{n(n-1)}{2!}y^2 + \cdots, here y=2xy = 2x, n=2n = -2.

First term: 11.

Second term: ny=(2)(2x)=4xny = (-2)(2x) = -4x.

Third term: n(n1)2!y2=(2)(3)2(2x)2=3×4x2=12x2\dfrac{n(n-1)}{2!}y^2 = \dfrac{(-2)(-3)}{2}(2x)^2 = 3 \times 4x^2 = 12x^2.

So (1+2x)214x+12x2(1 + 2x)^{-2} \approx 1 - 4x + 12x^2, valid for 2x<1|2x| < 1, that is x<12|x| < \dfrac{1}{2}.

Markers reward the general-index expansion, correct first three terms, and the validity condition x<12|x| < \tfrac{1}{2}. Omitting the validity range loses a mark for this type of expansion.

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