Solved Examples on Recursive Sequences

Formulas: Formulas

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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.
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(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} $

(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 $

(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 $