Sequences and series flashcards

Test yourself one card at a time: say the answer, flip, then mark it. Cards you know drop out; the rest come back until you know them. Your progress stays in this browser.

Recurrence relation
A rule giving each term from the previous one, u(n+1) = f(u(n)); it needs a first term to define a sequence.
Increasing sequence
u(n+1) > u(n) for every n ≥ 1.
Decreasing sequence
u(n+1) < u(n) for every n ≥ 1.
Periodic sequence of order p
The terms repeat every p terms: u(n+p) = u(n) for all n, with p the smallest positive integer that works.
Arithmetic sequence
Each term is the previous term plus a fixed common difference d.
Geometric sequence
Each term is the previous term multiplied by a fixed non-zero common ratio r.
Convergent geometric series
One whose partial sums approach a finite limit, the sum to infinity; this happens exactly when |r| < 1.
nth term of an arithmetic sequence
u(n) = a + (n − 1)d (Yes)
Sum of the first n terms of an arithmetic series
S(n) = ½n[2a + (n − 1)d] = ½n(a + l), l = last term (Yes)
Sum of the first n natural numbers
1 + 2 + … + n = ½n(n + 1) (No — derive it)
nth term of a geometric sequence
u(n) = ar^(n − 1) (Yes)
Sum of the first n terms of a geometric series
S(n) = a(1 − rⁿ)/(1 − r) = a(rⁿ − 1)/(r − 1), r ≠ 1 (Yes)
Sum to infinity of a geometric series
S∞ = a/(1 − r), valid only for |r| < 1 (Yes)
Plotting u(n) against n: what shape do arithmetic terms make?
Points on a straight line with gradient d (only whole-number n are plotted).
Plotting u(n) against n: what shape do geometric terms make (a > 0, r > 1)?
Points on an exponential growth curve; equal ratios, not equal steps.
Partial sums S(n) of a geometric series with −1 < r < 0
They jump above and below S∞, getting closer each time.
Staircase diagram for u(n+1) = f(u(n))
Go up to y = f(x) for the next term, across to y = x to use it as the next input; a staircase closing in on the crossing point shows the terms approaching a limit.
Reverse-and-add picture for 2 + 5 + 8 + 11 + 14 + 17
Stack each term on its partner from the reversed list: every column is 19 = first + last, so 2S = 6 × 19 and S = 57.
Outline the proof of the arithmetic series formula
Write S(n) forwards and backwards; add term by term to get n pairs each equal to 2a + (n − 1)d; divide by 2.
Outline the proof of the geometric series formula
Write S(n), then rS(n) shifted one place; subtract so all but a and arⁿ cancel; divide by 1 − r.
Why is there no sum to infinity when |r| ≥ 1?
The terms do not shrink towards 0, so the partial sums grow without limit or keep oscillating.
How do you prove a sequence is increasing?
Show u(n+1) − u(n) > 0 for every n (e.g. simplify it to an expression that is clearly positive).
Conditions for x, y, z to be consecutive terms of an AP and of a GP
AP: 2y = x + z. GP: y² = xz.
u(5) = 23 and u(12) = 51 in an AP. First term above 200?
7d = 28, d = 4, a = 7. 7 + 4(n − 1) > 200 gives n > 49.25, so the 50th term (203).
Evaluate Σ (5r − 2) from r = 1 to 40
5 × ½ × 40 × 41 − 2 × 40 = 4100 − 80 = 4020.
Evaluate Σ (3r + 1) from r = 11 to 30
20 terms, first 34, last 91: ½ × 20 × 125 = 1250.
GP with u(2) = 60 and u(5) = 3.84. Find r, a and S∞.
r³ = 3.84/60 = 0.064, r = 0.4; a = 150; S∞ = 150/0.6 = 250.
Sum to infinity of 81 − 27 + 9 − …
r = −1/3, |r| < 1, S∞ = 81/(4/3) = 60.75.
u(n+1) = (u(n) + 3)/(u(n) − 1), u(1) = 2. Sum of the first 25 terms?
Terms 2, 5, 2, 5, … (period 2): 13 twos and 12 fives, 26 + 60 = 86.
Output 24 000 in year 1, rising 5% a year. First year above 50 000?
1.05^(n−1) > 25/12, n − 1 > log(25/12)/log 1.05 = 15.04, so year 17.
Smallest integer n with 0.9ⁿ < 0.01
n log 0.9 < log 0.01; log 0.9 < 0 so the inequality flips: n > 43.7, n = 44.
Term or sum?
“Amount in year n” is a term u(n); “total over n years” is a sum S(n). Decide before choosing a formula.
Dividing an inequality by log r when 0 < r < 1
log r is negative, so the inequality sign reverses.
Number of terms in Σ from r = p to r = q
q − p + 1 (both ends included).
Giving n from an inequality
n must be a positive integer: give the smallest integer above the critical value and check the neighbours if it is close to a whole number.
Using S∞ = a/(1 − r)
State |r| < 1 first; reject any value of r outside −1 < r < 1.
Summing many terms of a periodic sequence
Count complete cycles and the leftover terms separately; never multiply by a fractional number of cycles.

Practise this topicNotes Questions (28)Test builder