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Sequences and sigma notation questions
A sequence is a list with a law attached, given either as a formula for the nth term or as a recurrence that builds each term from the one before. Three words cover how such lists behave. One Greek letter compresses any sum of their terms into a single readable instruction.
7 original questions · 25 marks · the sequences and sigma notation notes · Sequences and series
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A sequence has nth term un = 2n2 − 1. Write down the first three terms, and determine whether 71 is a term of the sequence.
Worked answer
The terms are 1, 7, 17. For 71: 2n2 − 1 = 71 gives n2 = 36, so n = 6, a positive whole number: yes, 71 is the sixth term. B1 for the three terms, M1 for setting 2n2 − 1 = 71 and A1 for n = 6 with the conclusion. Had n come out fractional, the answer would be no, and showing the failed equation would be the proof.A sequence has u1 = 2 and un+1 = 2un + 1. Find the first four terms.
Worked answer
u2 = 2 × 2 + 1 = 5, then u3 = 2 × 5 + 1 = 11, then u4 = 2 × 11 + 1 = 23. M1 for a correct substitution into the recurrence, A1 for all three further terms. Show each feed of the machine. Recurrence questions are marked on the substitutions, so a bare list of four numbers can score less than working that contains one arithmetic slip.A sequence has u1 = 4 and un+1 = 12/un. Find the first four terms, and describe the sequence's behaviour.
Worked answer
The terms run 4, 3, 4, 3, so the sequence is periodic with period 2 and bounces between two values for ever. Dividing 12 by either value returns the other, so the pair traps itself. M1 for a correct division, A1 for the four terms, B1 for periodic with period 2. The word periodic and the number 2 are both wanted; describing it as “repeating” without the period is worth less.Evaluate Σ(3r + 2) for r running from 1 to 5.
Worked answer
The recipe gives 5, 8, 11, 14, 17, and the Σ says add them, so the total is 55. Splitting the sum works just as well: 3(1 + 2 + 3 + 4 + 5) + 2 × 5 = 45 + 10 = 55, since the constant 2 is added five times over. M1 for generating the terms or splitting the sum, A1 for the five values, A1 for 55. Treating Σ2 as 2 rather than 10 is the standard error, and it is the reason the second route needs care.A sequence has un = 50 − 4n. Find u12, determine which term is the first negative one, and evaluate the sum of the first five terms.
Worked answer
u12 = 50 − 48 = 2. A negative term needs 50 − 4n < 0, so n > 12.5, and since n counts terms the first negative one is the 13th, u13 = −2. The first five terms are 46, 42, 38, 34 and 30, which sum to 190. B1 for u12 = 2, M1 for 50 − 4n < 0, A1 for the 13th term, M1 for the five terms, A1 for 190. Rounding 12.5 down to 12 is the trap: the inequality has to be satisfied, so n climbs to the next whole number.A sequence has un = 2n/(n + 1). Show that the sequence is increasing, and that every term is less than 2.
Worked answer
un+1 − un = 2(n + 1)/(n + 2) − 2n/(n + 1) = 2[(n + 1)2 − n(n + 2)]/[(n + 1)(n + 2)] = 2/[(n + 1)(n + 2)], which is positive for every n: increasing. Also un = 2 − 2/(n + 1) < 2, since something positive is always subtracted. M1 for forming un+1 − un, A1 for the simplified 2/[(n + 1)(n + 2)], A1 for stating it is positive, M1 for un = 2 − 2/(n + 1), A1 for the conclusion. The sequence climbs for ever yet never reaches 2: increasing does not mean unbounded.Evaluate Σ(2r − 1) for r running from 1 to 20, and explain what the answer suggests about sums of odd numbers.
Worked answer
Split the sum: 2(1 + 2 + … + 20) − (1 + 1 + … + 1) = 2 × 210 − 20 = 400, using Σ1 = 20 for the constant part. And 400 = 202: the first n odd numbers always sum to n2, each new odd number wrapping another layer round a square. M1 for splitting the sum, A1 for 2 × 210 − 20, A1 for 400, B1 for the link with n2.
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