| Type of rule | What it gives you | Example |
|---|---|---|
| $n$th term (position-to-term) | Any term directly from its position $n$. | $u_n = 3n^2 - 2$ |
| Recurrence relation (term-to-term) | Each term from the one before, plus a starting value. | $u_{n+1} = 2u_n - 3,\; u_1 = 5$ |
A sequence has $n$th term $u_n = 3n^2 - 2$. Write down the first three terms, and find which term is equal to $145$.
The same rule with a different starting value gives a completely different sequence, so the first term is part of the definition. Some textbooks write $x_{n+1} = \mathrm{f}(x_n)$; the letter does not matter.
$u_{n+1} = 2u_n - 3$ with $u_1 = 5$. Find $u_2$, $u_3$ and $u_4$.
A sequence is defined by $u_{n+1} = k u_n + 4$, $u_1 = 2$. Given that $u_3 = 20$, find the possible values of $k$.
| Word | Definition | Example |
|---|---|---|
| Increasing | $u_{n+1} > u_n$ for all $n$ | $2, 5, 8, 11, \ldots$ |
| Decreasing | $u_{n+1} < u_n$ for all $n$ | $1, \tfrac12, \tfrac14, \tfrac18, \ldots$ |
| Periodic | The terms repeat: $u_{n+k} = u_n$ for all $n$ | $2, -1, \tfrac12, 2, -1, \tfrac12, \ldots$ |
The smallest such $k$ is the order (or period) of a periodic sequence. A sequence can also be none of these, for example $1, -2, 3, -4, \ldots$ (it oscillates but never repeats).
Prove that the sequence with $n$th term $u_n = \dfrac{n}{n+1}$ is increasing.
$u_{n+1} = \dfrac{1}{1 - u_n}$ with $u_1 = 2$. Show that the sequence is periodic, state its order, and find $\displaystyle\sum_{r=1}^{100} u_r$.
$\Sigma$ (capital sigma) means "add up". The letter $r$ is a dummy variable: it takes every whole-number value from the bottom limit to the top limit, inclusive.
Evaluate $\displaystyle\sum_{r=1}^{5} (2r + 1)$.
The formula for $\sum r$ is proved on the arithmetic series page.
Evaluate $\displaystyle\sum_{r=1}^{20} (3r - 2)$.
Evaluate $\displaystyle\sum_{r=11}^{30} r$.
Write $4 + 7 + 10 + \cdots + 61$ using sigma notation.
$n$th term
Position to term: substitute $n$ directly.
Recurrence
$u_{n+1} = \mathrm{f}(u_n)$ plus a first term.
Is it a term?
Solve $u_n = $ value; it is a term only if $n$ is a positive integer.
Increasing
Prove $u_{n+1} - u_n > 0$ for all $n$.
Periodic
$u_{n+k} = u_n$; order $k$ is the shortest repeat.
Limit
If it converges, $L = \mathrm{f}(L)$.
Counting terms
$r = a$ to $r = b$ has $b - a + 1$ terms.
Standard sums
$\sum 1 = n$, $\sum r = \tfrac12 n(n+1)$.
Not from 1
$\sum_{a}^{b} = \sum_{1}^{b} - \sum_{1}^{a-1}$.
A sequence has $n$th term $u_n = 5n - 3$. Find $u_1$ and $u_{10}$, and find which term equals $92$.
▶ Show solution
$u_1 = 5 - 3 = 2$ and $u_{10} = 50 - 3 = 47$.
$5n - 3 = 92 \Rightarrow 5n = 95 \Rightarrow n = 19$. So $92$ is the $19$th term.
$u_{n+1} = 3u_n - 1$ with $u_1 = 2$. Find $u_2$, $u_3$ and $u_4$.
▶ Show solution
$u_2 = 3(2) - 1 = 5$
$u_3 = 3(5) - 1 = 14$
$u_4 = 3(14) - 1 = 41$
Write $2 + 5 + 8 + \cdots + 59$ in sigma notation.
▶ Show solution
The terms increase by $3$ from $2$, so $u_r = 3r - 1$.
$3r - 1 = 59 \Rightarrow r = 20$.
$\displaystyle 2 + 5 + \cdots + 59 = \sum_{r=1}^{20} (3r - 1)$
Evaluate $\displaystyle\sum_{r=1}^{4} 2^r$ and $\displaystyle\sum_{r=0}^{3} (r^2 + 1)$.
▶ Show solution
$\displaystyle\sum_{r=1}^{4} 2^r = 2 + 4 + 8 + 16 = 30$
$\displaystyle\sum_{r=0}^{3} (r^2 + 1) = 1 + 2 + 5 + 10 = 18$ (note that $r = 0$ is included, giving four terms).
Is the sequence $u_n = n^2 - 6n$ increasing, decreasing, or neither? Justify your answer.
▶ Show solution
$u_{n+1} - u_n = (n+1)^2 - 6(n+1) - n^2 + 6n = 2n - 5$.
This is negative for $n = 1, 2$ and positive for $n \geqslant 3$.
The terms are $-5, -8, -9, -8, -5, \ldots$: they fall then rise, so the sequence is neither increasing nor decreasing.
Find $a$ and $b$ given that the sequence $u_{n+1} = a u_n + b$ has $u_1 = 4$, $u_2 = 7$ and $u_3 = 13$. Hence find $u_4$.
▶ Show solution
$u_2$: $4a + b = 7$. $u_3$: $7a + b = 13$.
Subtract: $3a = 6$, so $a = 2$ and $b = 7 - 8 = -1$.
$u_4 = 2(13) - 1 = 25$.
$u_{n+1} = -\dfrac{1}{u_n}$ with $u_1 = 3$. Show that the sequence is periodic, state its order, and find $\displaystyle\sum_{r=1}^{25} u_r$.
▶ Show solution
$u_2 = -\tfrac13$, $u_3 = -\dfrac{1}{-\frac13} = 3 = u_1$. The sequence is periodic with order $2$.
Each pair sums to $3 - \tfrac13 = \tfrac83$.
$25 = 2 \times 12 + 1$: twelve pairs, then $u_{25} = u_1 = 3$.
$\displaystyle\sum_{r=1}^{25} u_r = 12 \times \tfrac83 + 3 = 32 + 3 = 35$
Evaluate $\displaystyle\sum_{r=10}^{25} (2r - 1)$.
▶ Show solution
$\displaystyle\sum_{r=1}^{n} (2r - 1) = 2 \times \tfrac12 n(n+1) - n = n^2$.
$\displaystyle\sum_{r=10}^{25} (2r-1) = \sum_{r=1}^{25} (2r-1) - \sum_{r=1}^{9} (2r-1) = 625 - 81 = 544$
Show that $\displaystyle\sum_{r=1}^{n} (4r + 3) = 2n^2 + 5n$, and hence find $n$ such that $\displaystyle\sum_{r=1}^{n} (4r + 3) = 250$.
▶ Show solution
$\displaystyle\sum_{r=1}^{n} (4r + 3) = 4 \times \tfrac12 n(n+1) + 3n = 2n^2 + 2n + 3n = 2n^2 + 5n$ ✓
$2n^2 + 5n - 250 = 0 \Rightarrow (2n + 25)(n - 10) = 0$.
$n$ must be a positive integer, so $n = 10$.
A lake is stocked with fish. Each year $20\%$ of the fish are caught, and then $300$ new fish are added. At the start there are $1000$ fish. Let $P_n$ be the number of fish after $n$ years.
(a) Explain why $P_{n+1} = 0.8P_n + 300$, with $P_0 = 1000$. (b) Find $P_1$, $P_2$ and $P_3$. (c) Given that the sequence converges, find its limit. (d) Verify that $P_n = 1500 - 500(0.8)^n$ satisfies the recurrence relation, and hence find the first year in which the population exceeds $1450$.
▶ Show solution
(a) Losing $20\%$ leaves $0.8P_n$; adding $300$ gives $0.8P_n + 300$.
(b) $P_1 = 800 + 300 = 1100$, $P_2 = 880 + 300 = 1180$, $P_3 = 944 + 300 = 1244$.
(c) $L = 0.8L + 300 \Rightarrow 0.2L = 300 \Rightarrow L = 1500$ fish.
(d) $P_0 = 1500 - 500 = 1000$ ✓. $0.8P_n + 300 = 1200 - 400(0.8)^n + 300 = 1500 - 500(0.8)^{n+1} = P_{n+1}$ ✓
Need $1500 - 500(0.8)^n > 1450$, i.e. $(0.8)^n < 0.1$.
Take logs: $n \ln 0.8 < \ln 0.1$. Since $\ln 0.8 < 0$ the inequality reverses: $n > \dfrac{\ln 0.1}{\ln 0.8} = 10.32\ldots$
So $n = 11$: the population first exceeds $1450$ after $11$ years ($P_{10} \approx 1446$, $P_{11} \approx 1457$).