If there is one prayer that you should pray/sing every day and every hour, it is the LORD's prayer (Our FATHER in Heaven prayer)
It is the most powerful prayer. A pure heart, a clean mind, and a clear conscience is necessary for it.
- Samuel Dominic Chukwuemeka

For in GOD we live, and move, and have our being. - Acts 17:28

The Joy of a Teacher is the Success of his Students. - Samuel Chukwuemeka

Solved Examples on Recursive Sequences

Samuel Dominic Chukwuemeka (SamDom For Peace) Prerequisites:
(1.) Linear Systems
(2.) Exponents
(3.) Quadratic Equations
(4.) Quadratic Functions

Formulas: Formulas
Calculators: Calculators

For ATAR Students
The calculator free questions are indicated accordingly. Those questions are solved in a way that does not require a calculator.

For ACT Students
The ACT is a timed exam...$60$ questions for $60$ minutes
This implies that you have to solve each question in one minute.
Some questions will typically take less than a minute a solve.
Some questions will typically take more than a minute to solve.
The goal is to maximize your time. You use the time saved on those questions you solved in less than a minute, to solve the questions that will take more than a minute.
So, you should try to solve each question correctly and timely.
So, it is not just solving a question correctly, but solving it correctly on time.
Please ensure you attempt all ACT questions.
There is no negative penalty for a wrong answer.

For JAMB and CMAT Students
Calculators are not allowed. So, the questions are solved in a way that does not require a calculator.

For WASSCE Students
Any question labeled WASCCE is a question for the WASCCE General Mathematics
Any question labeled WASSCE-FM is a question for the WASSCE Further Mathematics/Elective Mathematics

For GCSE Students
All work is shown to satisfy (and actually exceed) the minimum for awarding method marks.
Calculators are allowed for some questions. Calculators are not allowed for some questions.

For NSC Students
For the Questions:
Any space included in a number indicates a comma used to separate digits...separating multiples of three digits from behind.
Any comma included in a number indicates a decimal point.
For the Solutions:
Decimals are used appropriately rather than commas
Commas are used to separate digits appropriately.

Solve all questions
Use at least two methods whenever applicable.
Show all work

(1.)


$ nth\:\:term = 5 + \dfrac{2}{3^{n - 2}} \\[5ex] 4th\:\:term = 5 + \dfrac{2}{3^{4 - 2}} \\[5ex] = 5 + \dfrac{2}{3^2} \\[5ex] = 5 + \dfrac{2}{9} \\[5ex] = \dfrac{45}{9} + \dfrac{2}{9} \\[5ex] = \dfrac{45 + 2}{9} \\[5ex] = \dfrac{47}{9} \\[5ex] 5th\:\:term = 5 + \dfrac{2}{3^{5 - 2}} \\[5ex] = 5 + \dfrac{2}{3^3} \\[5ex] = 5 + \dfrac{2}{27} \\[5ex] = \dfrac{135}{27} + \dfrac{2}{27} \\[5ex] = \dfrac{135 + 2}{27} \\[5ex] = \dfrac{137}{27} \\[5ex] 4th\:\:term + 5th\:\:term \\[3ex] = \dfrac{47}{9} + \dfrac{137}{27} \\[5ex] = \dfrac{141}{27} + \dfrac{137}{27} \\[5ex] = \dfrac{141 + 137}{27} \\[5ex] = \dfrac{278}{27} $
(2.) ACT A sequence is defined for all positive integers by $s_n = 2s_{(n - 1)} + n + 1$ and $s_1 = 3$
What is $s_4?$

$ F.\:\: 9 \\[3ex] G.\:\: 18 \\[3ex] H.\:\: 22 \\[3ex] J.\:\: 49 \\[3ex] K.\:\: 111 \\[3ex] $

Recursive Seqeunce

$ s_n = 2s_{(n - 1)} + n + 1 \\[3ex] s_1 = 3 \\[3ex] s_2 = 2s_{(2 - 1)} + 2 + 1 \\[3ex] s_2 = 2s_1 + 2 + 1 \\[3ex] s_2 = 2(3) + 2 + 1 \\[3ex] s_2 = 6 + 2 + 1 \\[3ex] s_2 = 9 \\[3ex] s_3 = 2s_{(3 - 1)} + 3 + 1 \\[3ex] s_3 = 2s_2 + 3 + 1 \\[3ex] s_3 = 2(9) + 3 + 1 \\[3ex] s_3 = 18 + 3 + 1 \\[3ex] s_3 = 22 \\[3ex] s_4 = 2s_{(4 - 1)} + 4 + 1 \\[3ex] s_4 = 2s_3 + 4 + 1 \\[3ex] s_4 = 2(22) + 4 + 1 \\[3ex] s_4 = 44 + 4 + 1 \\[3ex] s_4 = 49 $
(3.) ACT The first $5$ terms of a sequence are given in the table below.
The sequence is defined by setting $a_1 = 9$ and $a_n = a_{n - 1} + (n - 1)^2$ for $n \ge 2$
What is the sixth term, $a_6$, of this sequence?

$a_1$ $a_2$ $a_3$ $a_4$ $a_5$ $a_6$
$9$ $10$ $14$ $23$ $39$ $?$

$ A.\:\: 62 \\[3ex] B.\:\: 64 \\[3ex] C.\:\: 76 \\[3ex] D.\:\: 78 \\[3ex] E.\:\: 95 \\[3ex] $

Recursive Seqeunce

$ a_n = a_{n - 1} + (n - 1)^2 \\[3ex] a_6 = a_{6 - 1} + (6 - 1)^2 \\[3ex] a_6 = a_{5} + 5^2 \\[3ex] a_6 = 39 + 25 \\[3ex] a_6 = 64 $
(4.) ACT The recursive formula for a sequence is given below, where $a_n$ is the value of the nth term.

$ a_1 = 10 \\[3ex] a_n = a_{n - 1} + 5 \\[3ex] $ Which of the following equations is an explicit formula for this sequence?

$ A.\:\: a_n = -5n + 10 \\[3ex] B.\:\: a_n = 5n + 5 \\[3ex] C.\:\: a_n = 5n + 10 \\[3ex] D.\:\: a_n = 10n - 5 \\[3ex] E.\:\: a_n = 10n + 5 \\[3ex] $

Recursive Seqeunce

$ a_1 = 10 \\[3ex] a_n = a_{n - 1} + 5 \\[3ex] a_2 = a_{2 - 1} + 5 \\[3ex] a_2 = a_1 + 5 = 10 + 5 = 15 \\[3ex] a_3 = a_{3 - 1} + 5 \\[3ex] a_3 = a_2 + 5 = 15 + 5 = 20 \\[3ex] a_1, a_2, a_3 = 10, 15, 20...arithmetic\:\:sequence \\[3ex] a = a_1 = 10 \\[3ex] d = 15 - 10 = 5 \\[3ex] a_n = a + d(n - 1) \\[3ex] a_n = 10 + 5(n - 1) \\[3ex] a_n = 10 + 5n - 5 \\[3ex] a_n = 5n + 5 $
(5.)


$ 1st\:\:term = a \\[3ex] 2nd\:\:term = (a - 1) * 5 \\[3ex] = 5(a - 1) \\[3ex] = 5a - 5 \\[3ex] 3rd\:\:term = [(5a - 5) - 1] * 5 \\[3ex] = [5a - 5 - 1] * 5 \\[3ex] = (5a - 6) * 5 \\[3ex] = 5(5a - 6) \\[3ex] = 25a - 30 \\[3ex] 3rd\:\:term = 120 \\[3ex] \rightarrow 25a - 30 = 120 \\[3ex] 25a = 120 + 30 \\[3ex] 25a = 150 \\[3ex] a = \dfrac{150}{25} \\[5ex] a = 6 \\[3ex] $ The first term is $6$
(6.) GCSE A sequence of numbers is formed by the iterative process

$ u_{n + 1} = \dfrac{3}{u_n + 1},\:\:\:\:\:\:\: u_1 = 4 \\[5ex] $ Work out the values of $u_2$ and $u_3$


Recursive Seqeunce

$ u_1 = 4 \\[3ex] u_{n + 1} = \dfrac{3}{u_n + 1} \\[5ex] For\:\: n = 1 \\[3ex] u_{1 + 1} = \dfrac{3}{u_1 + 1} \\[5ex] u_2 = \dfrac{3}{4 + 1} \\[5ex] u_2 = \dfrac{3}{5} \\[5ex] For\:\: n = 2 \\[3ex] u_{2 + 1} = \dfrac{3}{u_2 + 1} \\[5ex] u_3 = \dfrac{3}{\dfrac{3}{5} + 1} \\[7ex] \dfrac{3}{5} + 1 = \dfrac{3}{5} + \dfrac{5}{5} = \dfrac{3 + 5}{5} = \dfrac{8}{5} \\[5ex] u_3 = \dfrac{3}{\dfrac{8}{5}} \\[7ex] = 3 \div \dfrac{8}{5} \\[5ex] = 3 * \dfrac{5}{8} \\[5ex] u_3 = \dfrac{15}{8} $
(7.)

(8.) NSC The first FOUR terms of the sequence of numbers are $2;\;\;5;\;\;10$ and $17$
(8.1) Write down the next TWO terms in the sequence.
(8.2) Write down a recursive formula for the sequence.


$ 2,\;\;5,\;\;10,\;\;17 \\[3ex] RS_{1} = 2 \\[3ex] RS_{2} = 5 \\[3ex] RS_{3} = 10 \\[3ex] RS_4 = 17 \\[3ex] 5 - 2 = 3 \\[3ex] 10 - 5 = 5 \\[3ex] 17 - 10 = 7 \\[3ex] 3,\;\;5,\;\;7,\;\;9,\;\;11 \\[3ex] (8.1) \\[3ex] RS_5 = 17 + 9 = 26 \\[3ex] RS_6 = 26 + 11 = 37 \\[3ex] (8.2) \\[3ex] RS_{1} = 2 \\[3ex] RS_{2} = 5 \\[3ex] = 2 + 2 + 1 \\[3ex] = RS_{1} + 2(1) + 1 \\[3ex] RS_{3} = 10 \\[3ex] = 5 + 4 + 1 \\[3ex] = RS_{2} + 2(2) + 1 \\[3ex] RS_{4} = 17 \\[3ex] = 10 + 6 + 1 \\[3ex] = RS_{3} + 2(3) + 1 \\[3ex] RS_{5} = 26 \\[3ex] = 17 + 8 + 1 \\[3ex] = RS_4 + 2(4) + 1 \\[3ex] \therefore RS_{n + 1} \\[3ex] = RS_{n} + 2(n) + 1 \\[3ex] RS_{n + 1} = RS_{n} + 2n + 1 \;\;where\;\;n \ge 1 $
(9.)

$ 1st\:\:term = a = \dfrac{4}{2} = 2 \\[5ex] Test\:\:each\:\:option \\[3ex] Option\:\:A \\[3ex] nth\:\:term = \dfrac{3n + 1}{n + 1} \\[5ex] For\:\: n = 1, \implies a = 2 \\[3ex] nth\:\:term = \dfrac{3(1) + 1}{1 + 1} \\[5ex] = \dfrac{3 + 1}{2} \\[5ex] = \dfrac{4}{2} \\[5ex] = 2 \checkmark \\[3ex] For\:\: n = 2, \implies second\:\:term = \dfrac{7}{3} \\[5ex] nth\:\:term = \dfrac{3(2) + 1}{2 + 1} \\[5ex] = \dfrac{6 + 1}{3} \\[5ex] = \dfrac{7}{3} \checkmark \\[5ex] For\:\: n = 3, \implies third\:\:term = \dfrac{10}{4} \\[5ex] nth\:\:term = \dfrac{3(3) + 1}{3 + 1} \\[5ex] = \dfrac{9 + 1}{4} \\[5ex] = \dfrac{10}{4} \checkmark \\[5ex] For\:\: n = 4, \implies fourth\:\:term = \dfrac{13}{5} \\[5ex] nth\:\:term = \dfrac{3(4) + 1}{4 + 1} \\[5ex] = \dfrac{12 + 1}{5} \\[5ex] = \dfrac{13}{5} \checkmark \\[5ex] Option\:\:A\:\:is\:\:the\:\:correct\:\:option $
(10.)

(11.)

(12.)

$ Q_n = 3 * 2^{n - 2} \\[3ex] Q_2 = 3 * 2^{2 - 2} \\[3ex] Q_2 = 3 * 2^0 \\[3ex] Q_2 = 3 * 1 \\[3ex] Q_2 = 3 \\[3ex] U_m = 3 * 2^{2m - 3} \\[3ex] U_2 = 3 * 2^{2(2) - 3} \\[3ex] U_2 = 3 * 2^{4 - 3} \\[3ex] U_2 = 3 * 2^1 \\[3ex] U_2 = 3 * 2 \\[3ex] U_2 = 6 \\[3ex] Q_2 * U_2 = 3(6) = 18 $
(13.) ACT

Notice the pattern

$ 1 + 3\:\:are\:\:2\:\:numbers \\[3ex] 1 + 3 = 4 = 2^2 \\[3ex] 1 + 3 + 5\:\:are\:\:3\:\:numbers \\[3ex] 1 + 3 + 5 = 9 = 3^3 \\[3ex] 1 + 3 + 5 + 7 \:\:are\:\:4\:\:numbers \\[3ex] 1 + 3 + 5 + 7 = 16 = 4^2 \\[3ex] 1 + 3 + 5 + 7 + 9 \:\:are\:\:5\:\:numbers \\[3ex] 1 + 3 + 5 + 7 + 9 = 5^2 \\[3ex] Similarly \\[3ex] For\:\:144 \\[3ex] 144 = 12^2 \\[3ex] We\:\:need\:\:12\:\:numbers \\[3ex] 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 \:\:are\:\:12\:\:numbers \\[3ex] 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 = 12^2 \\[3ex] Highest\:\:odd\:\:number = 23 $
(14.)

Triangular Seqeunce

$ TS_n = \dfrac{n(n + 1)}{2} \\[5ex] TS_7 = \dfrac{7(7 + 1)}{2} \\[5ex] TS_7 = \dfrac{7 * 8}{2} \\[5ex] TS_7 = 7(4) \\[3ex] TS_7 = 28 $