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Sequences and Series: Arithmetic, Geometric and the Binomial Expansion

Sequences build term by term, via a direct nth-term rule or a recurrence xn+1 = f(xn); a series adds the terms. Arithmetic and geometric progressions have exact sum formulas, convergent GPs have a sum to infinity, and (1+x)n expands with binomial coefficients.

Part of the TMUA syllabus — revision for the Test of Mathematics for University Admission (TMUA), the UAT-UK maths test used by Cambridge, Oxford, Imperial, UCL, LSE, Warwick and Durham. No calculator; multiple choice.

Key points

  • A sequence can be given by a closed nth-term formula (position → value directly) or by a recurrence xn+1 = f(xn); a recurrence must be iterated step by step from a stated starting value.
  • Arithmetic progression (AP): constant difference d, nth term a + (n-1)d, and the special sum 1 + 2 +⋯ + n = n(n+1)/2.
    the case a = d = 1
  • Geometric progression (GP): constant ratio r, nth term a × rn-1, finite sum Sn = a(1 - rn)/(1 - r) for r ≠ 1.
  • A geometric series has a sum to infinity a/(1 - r) ONLY when |r| < 1; otherwise the partial sums grow without bound (the series diverges).
  • For positive integer n the expansion of (1+x)n is finite with n+1 terms, and the coefficients are nCr = n!/(r!(n-r)!) (the rows of Pascal's triangle).
  • In (a + f(x))n the term for power r is nCr × an-r × (f(x))^r; the power of a and the power of f(x) always add up to n.
  • n! = n(n-1)⋯2*1 with 0! = 1, and nCr = nC(n-r) (symmetry) lets you compute the easier of two equal coefficients.
  • Three numbers are in GP when the middle squared equals the product of the outers; in an AP consecutive terms differ by the same d.
    b2 = a × c, the geometric-mean check

Diagram

term ntotal123456122436sum to infinity = 36
A convergent geometric series (first term 24, ratio 1/3): the terms shrink and the running total approaches the sum to infinity, 36.

Formulae

un = a + (n-1)d

nth term of an arithmetic progression with first term a and common difference d.

Sn = (n/2)(2a + (n-1)d)

Sum of n terms of an AP; also equals (n/2)(a + l) with last term l. The sum 1 + 2 + ⋯ + n = n(n+1)/2 is the case a = d = 1.

un = a × rn-1

nth term of a geometric progression with first term a and common ratio r (note the power is n-1, not n).

Sn = a(1 - rn)/(1 - r)

Sum of the first n terms of a GP (r ≠ 1); equivalently a(rn - 1)/(r - 1) when r > 1.

Sinf = a/(1 - r)

Sum to infinity of a convergent GP; valid ONLY when |r| < 1, otherwise no finite sum exists.

nCr = n!/(r!(n-r)!)

Binomial coefficient. The term in x^r of (1+x)n is nCr × x^r, so (1+x)n = 1 + nx + n(n-1)/2 × x2 + ⋯ for positive integer n.

Definitions

Common difference (d)
The constant amount added to pass from each term to the next in an arithmetic sequence; it may be negative.
d = un+1 - un
Common ratio (r)
The constant multiplier between consecutive terms of a geometric sequence; the series converges only when |r| < 1.
r = un+1/un
Recurrence relation
A rule that generates each term from the previous one(s), e.g. xn+1 = f(xn), and needs a stated starting value x1 to define the whole sequence.
Convergent series
A series whose partial sums approach a finite limit; a geometric series converges exactly when |r| < 1, with sum to infinity a/(1 - r).

Worked examples

1

A geometric series has first term 24 and common ratio 1/3. Find (a) the fifth term, (b) the sum of the first four terms, and (c) the sum to infinity.

  1. 1

    Note the values:

    a = 24, r = 1/3
  2. 2

    Fifth term uses r4:

    u5 = 24 × (1/3)4 = 24/81 = 8/27
  3. 3

    Sum of four terms:

    S4 = 24(1 - (1/3)4)/(1 - 1/3)
  4. 4

    Simplify (divide by 2/3 means times 3/2):

    S4 = 24 × (80/81) × (3/2) = 320/9
  5. 5

    Since |1/3| < 1 the sum to infinity exists:

    Sinf = 24/(1 - 1/3) = 24 × (3/2) = 36

Answer: (a) u5 = 8/27; (b) S4 = 320/9; (c) Sinf = 36. Note S4 = 320/9 (about 35.6) is just below Sinf = 36, as expected.

2

Expand (2 + 3x)4 in ascending powers of x, and state the coefficient of x2.

  1. 1

    Set the parts:

    a = 2, b = 3x, n = 4
  2. 2

    Pascal row 4:

    1, 4, 6, 4, 1.

  3. 3

    x1 term:

    4 × 23 × (3x) = 4 × 8 × 3x = 96x.

  4. 4

    x2 term:

    6 × 22 × (3x)2 = 6 × 4 × 9x2 = 216x2.

  5. 5

    Remaining terms:

    24 = 16, then 4 × 2 × 27x3 = 216x3 and (3x)4 = 81x4

Answer: (2 + 3x)4 = 16 + 96x + 216x2 + 216x3 + 81x4; the coefficient of x2 is 216.

Common mistakes

  • ×Off-by-one in the power or term count: the nth term of a GP is a × rn-1 (not a × rn), and an AP uses (n-1)d, so the 5th term uses r4, not r5.
  • ×Applying the sum to infinity when |r| ≥ 1 - it only exists for a convergent series with |r| < 1; always check the ratio before using a/(1 - r).
  • ×Forgetting to raise the whole bracket: in (a + bx)n the term needs (bx)^r, so both b and x are raised to r; writing b × x^r drops a factor of br-1.
  • ×Confusing first term a with the difference d or ratio r, or miscounting how many terms are being summed when a range such as the 3rd to 10th term is requested.
  • ×Arithmetic slips when dividing by (1 - r) for fractional r: dividing by 2/3 means multiplying by 3/2, not by 2/3.

No-calculator tips

  • ✓Keep everything as exact fractions; dividing by (1 - r) is multiplying by its reciprocal (e.g. /(2/3) = *(3/2)), so you never need decimals.
  • ✓Read Pascal's triangle rows straight off for small n (n ≤ 6) instead of recomputing n!/(r!(n-r)!) each time, and use the symmetry nCr = nC(n-r) to save work.
  • ✓Eliminate options by size: a convergent GP's partial sums creep up towards a/(1 - r), so any 'sum of the first k terms' must be slightly below the sum to infinity - reject any option above it.
  • ✓For 'which term is negative' or 'when do terms fall below X', set up an inequality with a + (n-1)d or a × rn-1 rather than listing every term.
  • ✓Sanity-check a binomial coefficient: the powers in each term must add to n, and the first and last coefficients are always 1; a wrong middle number usually means a slipped factor of a or b.

Test yourself

Original practice questions, no calculator. Work each out before revealing the answer.

Q1.In the binomial expansion of (1 - 2x)6, what is the coefficient of x3?

  • A. 160
  • B. -60
  • C. -20
  • D. -160
  • E. -8
Show answer

Answer: D — -160

The term is C(6,3) × (-2x)3 = 20 × (-8) x3 = -160 x3. Common slips: dropping the minus sign (160), using 2*3 instead of 23 (-60), forgetting to cube the 2 (-20), or omitting the binomial coefficient (-8).

Q2.A geometric series has first term 18 and a sum to infinity of 12. What is the common ratio r?

  • A. r = 1/2
  • B. r = 1/3
  • C. r = 2/3
  • D. r = 3/2
  • E. r = -1/2
Show answer

Answer: E — r = -1/2

Sum to infinity a/(1 - r) = 12 with a = 18 gives 1 - r = 18/12 = 3/2, so r = -1/2 (and |r| < 1 confirms it converges). The value 3/2 ignores the |r| < 1 condition; 2/3 and 1/3 come from inverting the fraction; 1/2 is a sign slip.

Q3.In an arithmetic progression the 5th term is 17 and the 9th term is 33. What is the sum of the first 10 terms?

  • A. 210
  • B. 190
  • C. 150
  • D. 380
  • E. 170
Show answer

Answer: B — 190

Common difference d = (33 - 17)/(9 - 5) = 4, and a + 4d = 17 gives a = 1, so S10 = (10/2)(2*1 + 9*4) = 5 × 38 = 190. Distractors come from using n instead of n-1 (210), dropping the 1/2 factor (380), or an off-by-one in the term formula (150).

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