Questions on sequences and sigma notation

Question 1

A sequence has nth term u_n = 2n² − 1. Write down the first three terms, and determine whether 71 is a term of the sequence.

Accepted 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.

Question 2

A sequence has u₁ = 2 and u_n+1 = 2u_n + 1. Find the first four terms.

Accepted 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 substitution: a bare list of four numbers does not show the recurrence being applied.

Question 3

Write the sum 7 + 12 + 17 + 22 + 27 + 32 using sigma notation.

Accepted Answer

The terms climb in steps of 5 from 7, so term r is 5r + 2: at r = 1 that gives 7 and at r = 6 it gives 32. B1 for the general term 5r + 2. There are six terms, so the counter runs from 1 to 6 and the sum is Σ(5r + 2) for r = 1 to 6. B1 for the limits.

Question 4

Evaluate Σ(r² + 1) for r running from 3 to 7.

Accepted Answer

The counter starts at 3 rather than at 1, so the terms are the values of r² + 1 at r = 3, 4, 5, 6 and 7: 10, 17, 26, 37 and 50. M1 for reading the range as r = 3 to r = 7, A1 for the five terms. Adding them, the total is 140. A1 for 140. Five terms are wanted, not seven: the count is 7 − 3 + 1, because both ends are included. With this few terms, and a term that is not linear in r, listing and adding is quicker than any rearrangement.

Question 5

A sequence has u₁ = 4 and u_n+1 = 12/u_n. Find the first four terms, and describe the sequence's behaviour.

Accepted 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.

Question 6

Evaluate Σ(3r + 2) for r running from 1 to 5.

Accepted 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.

Question 7

A sequence has u_n = 50 − 4n. Find u₁₂, determine which term is the first negative one, and evaluate the sum of the first five terms.

Accepted 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.

Question 8

A sequence is defined by u₁ = 1 and u_n+1 = 2u_n − 3. Evaluate Σu_r for r running from 1 to 5.

Accepted Answer

Run the recurrence: u₂ = 2 × 1 − 3 = −1, then u₃ = 2 × (−1) − 3 = −5, then u₄ = 2 × (−5) − 3 = −13, then u₅ = 2 × (−13) − 3 = −29. M1 for a correct substitution, A1 for the five terms 1, −1, −5, −13, −29. The Σ asks for their total: 1 − 1 − 5 − 13 − 29 = −47. A1 for −47. A recurrence gives no formula for u_r, so every term has to be generated before the sum can be reached, and a slip in u₂ travels through all of them.

Question 9

A sequence is defined by u₁ = 2 and u_n+1 = ku_n + 3, where k is a constant. Given that u₃ = 23, find the two possible values of k.

Accepted Answer

Apply the recurrence twice, keeping k as a letter. u₂ = 2k + 3. M1 for u₂ = 2k + 3. Then u₃ = k(2k + 3) + 3 = 2k² + 3k + 3. M1 for u₃ in terms of k. Setting that equal to 23 gives 2k² + 3k − 20 = 0. A1 for the quadratic. It factorises as (2k − 5)(k + 4) = 0, so k = 5/2 or k = −4. A1 for both values. Both work: at k = 5/2 the terms run 2, 8, 23, and at k = −4 they run 2, −5, 23. Two applications of the recurrence square the unknown, which is what produces a second solution.

Question 10

A sequence has u_n = (−1)ⁿ(n + 2). Write down the first five terms, state whether the sequence is increasing, decreasing or neither, and evaluate Σu_r for r running from 1 to 10.

Accepted Answer

The factor (−1)ⁿ is −1 at odd n and +1 at even n, so the terms are −3, 4, −5, 6, −7. M1 for using the alternating factor, A1 for the five terms. The sequence is neither increasing nor decreasing: it rises from −3 to 4 and then falls from 4 to −5, so the comparison between consecutive terms does not point the same way at every step. B1 for neither, with the reason. For the sum, pair the ten terms as they come: (−3 + 4) + (−5 + 6) + (−7 + 8) + (−9 + 10) + (−11 + 12), five pairs each totalling 1, so the sum is 5. M1 for pairing consecutive terms, A1 for 5. Nor is the sequence periodic: the signs repeat every two terms, but the sizes grow without ever coming back, and a periodic sequence has to repeat its values.

Question 11

Given that Σ(2r + c) for r running from 1 to 15 is 300, find the value of the constant c.

Accepted Answer

Split the sum into its two parts, each with the counter running from 1 to 15: Σ(2r + c) = 2Σr + Σc. M1 for splitting the sum. The first part is 2(1 + 2 + … + 15) = 2 × 120 = 240. A1 for 240. The second adds the constant once for each of the fifteen terms, so Σc = 15c, which is Σ1 = n with n = 15. B1 for 15c. Then 240 + 15c = 300 gives 15c = 60 and c = 4. A1 for c = 4.

Question 12

A sequence has u_n = 2n/(n + 1). Show that the sequence is increasing, and that every term is less than 2.

Accepted Answer

u_n+1 − u_n = 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 u_n = 2 − 2/(n + 1) < 2, since something positive is always subtracted. M1 for forming u_n+1 − u_n, A1 for the simplified 2/[(n + 1)(n + 2)], A1 for stating it is positive, M1 for u_n = 2 − 2/(n + 1), A1 for the conclusion. The sequence climbs for ever yet never reaches 2: increasing does not mean unbounded.

Question 13

Evaluate Σ(2r − 1) for r running from 1 to 20, and explain what the answer suggests about sums of odd numbers.

Accepted 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.

Question 14

A sequence has u_n = n² − 8n + 3. Show that u_n+1 − u_n = 2n − 7, find the least term of the sequence, and show that the sequence increases from that term onwards.

Accepted Answer

Substituting n + 1, u_n+1 = (n + 1)² − 8(n + 1) + 3 = n² + 2n + 1 − 8n − 8 + 3 = n² − 6n − 4. M1 for u_n+1 expanded. Subtracting u_n = n² − 8n + 3 leaves 2n − 7. A1 for 2n − 7. That difference is −5, −3 and −1 at n = 1, 2 and 3, and +1 at n = 4, so u₂ < u₁, u₃ < u₂ and u₄ < u₃, while u₅ > u₄. M1 for the sign of 2n − 7, A1 for the change of sign between n = 3 and n = 4. The sequence therefore falls as far as u₄ and rises after it, so the least term is u₄ = 16 − 32 + 3 = −13. A1 for −13. For every n ≥ 4, 2n − 7 ≥ 1, which is positive, so u_n+1 > u_n and the sequence increases from u₄ onwards. B1 for the inequality holding for all n ≥ 4. The least term is unique: u₃ and u₅ are both −12, one on each side of −13, and since 2n − 7 is never zero for whole n, no two consecutive terms are ever equal.

Question 15

A sequence is defined by u₁ = a and u_n+1 = 5 − u_n, where a is a constant and a ≠ 2.5. Show that the sequence is periodic with order 2, and find, in terms of n, the sum of the first 2n terms.

Accepted Answer

Applying the recurrence, u₂ = 5 − a, and then u₃ = 5 − (5 − a) = a. M1 for u₂ and u₃. Since u₃ = u₁, and each term is fixed by the one before it, the whole list repeats: a, 5 − a, a, 5 − a, … A1 for the repetition, whatever a is. The two values are different because a ≠ 2.5, so the cycle is genuinely two long and the order is 2. B1 for the order with that reason. The first 2n terms are n copies of the pair a and 5 − a, and each pair totals a + (5 − a) = 5, so the sum is 5n. M1 for pairing the 2n terms into n pairs, A1 for 5n. At a = 2.5 the two values coincide and the sequence is constant, which is why that value is set aside; the sum 5n still holds there.