Exercise 2.4 - Solutions
1. Find the next three terms of the following sequences.
(i) \(8, 24, 72, \dots\)
Observe the pattern: Each term is obtained by multiplying the previous term by \(3\).
- \(8 \times 3 = 24\)
- \(24 \times 3 = 72\)
Calculating the next three terms:
- 4th term: \(72 \times 3 = 216\)
- 5th term: \(216 \times 3 = 648\)
- 6th term: \(648 \times 3 = 1944\)
Next three terms: 216, 648, 1944
(ii) \(5, 1, -3, \dots\)
Observe the pattern: Each term is obtained by subtracting \(4\) from the previous term.
- \(5 - 4 = 1\)
- \(1 - 4 = -3\)
Calculating the next three terms:
- 4th term: \(-3 - 4 = -7\)
- 5th term: \(-7 - 4 = -11\)
- 6th term: \(-11 - 4 = -15\)
Next three terms: -7, -11, -15
(iii) \(\dfrac{1}{4}, \dfrac{2}{9}, \dfrac{3}{16}, \dots\)
Observe the pattern of the \(n^{\text{th}}\) term:
- Numerators: \(1, 2, 3, \dots \implies n\)
- Denominators: \(4, 9, 16, \dots \implies (n+1)^2\)
Calculating the next three terms (\(n = 4, 5, 6\)):
- 4th term (\(n=4\)): \(\dfrac{4}{(4+1)^2} = \dfrac{4}{5^2} = \dfrac{4}{25}\)
- 5th term (\(n=5\)): \(\dfrac{5}{(5+1)^2} = \dfrac{5}{6^2} = \dfrac{5}{36}\)
- 6th term (\(n=6\)): \(\dfrac{6}{(6+1)^2} = \dfrac{6}{7^2} = \dfrac{6}{49}\)
Next three terms: \(\dfrac{4}{25}, \dfrac{5}{36}, \dfrac{6}{49}\)
2. Find the first four terms of the sequences whose \(n^{\text{th}}\) terms are given by:
(i) \(a_n = n^3 - 2\)
Substitute \(n = 1, 2, 3, 4\):
- \(a_1 = 1^3 - 2 = 1 - 2 = -1\)
- \(a_2 = 2^3 - 2 = 8 - 2 = 6\)
- \(a_3 = 3^3 - 2 = 27 - 2 = 25\)
- \(a_4 = 4^3 - 2 = 64 - 2 = 62\)
First four terms: -1, 6, 25, 62
(ii) \(a_n = (-1)^{n+1} n (n + 1)\)
Substitute \(n = 1, 2, 3, 4\):
- \(a_1 = (-1)^{1+1} \times 1 \times (1 + 1) = (-1)^2 \times 1 \times 2 = 1 \times 2 = 2\)
- \(a_2 = (-1)^{2+1} \times 2 \times (2 + 1) = (-1)^3 \times 2 \times 3 = -1 \times 6 = -6\)
- \(a_3 = (-1)^{3+1} \times 3 \times (3 + 1) = (-1)^4 \times 3 \times 4 = 1 \times 12 = 12\)
- \(a_4 = (-1)^{4+1} \times 4 \times (4 + 1) = (-1)^5 \times 4 \times 5 = -1 \times 20 = -20\)
First four terms: 2, -6, 12, -20
(iii) \(a_n = 2n^2 - 6\)
Substitute \(n = 1, 2, 3, 4\):
- \(a_1 = 2(1)^2 - 6 = 2(1) - 6 = -4\)
- \(a_2 = 2(2)^2 - 6 = 2(4) - 6 = 8 - 6 = 2\)
- \(a_3 = 2(3)^2 - 6 = 2(9) - 6 = 18 - 6 = 12\)
- \(a_4 = 2(4)^2 - 6 = 2(16) - 6 = 32 - 6 = 26\)
First four terms: -4, 2, 12, 26
3. Find the \(n^{\text{th}}\) term of the following sequences:
(i) \(2, 5, 10, 17, \dots\)
Observe the pattern:
- \(a_1 = 2 = 1^2 + 1\)
- \(a_2 = 5 = 2^2 + 1\)
- \(a_3 = 10 = 3^2 + 1\)
- \(a_4 = 17 = 4^2 + 1\)
\(a_n = n^2 + 1, \quad \forall n \in \mathbb{N}\)
(ii) \(0, \dfrac{1}{2}, \dfrac{2}{3}, \dots\)
Rewrite terms to identify the pattern:
- \(a_1 = 0 = \dfrac{1 - 1}{1} = \dfrac{0}{1}\)
- \(a_2 = \dfrac{1}{2} = \dfrac{2 - 1}{2}\)
- \(a_3 = \dfrac{2}{3} = \dfrac{3 - 1}{3}\)
\(a_n = \dfrac{n - 1}{n}, \quad \forall n \in \mathbb{N}\)
(iii) \(3, 8, 13, 18, \dots\)
This is an Arithmetic Progression (A.P.) where:
- First term \(a = 3\)
- Common difference \(d = 8 - 3 = 5\)
\(a_n = 5n - 2, \quad \forall n \in \mathbb{N}\)
4. Find the indicated terms of the sequences whose \(n^{\text{th}}\) terms are given by:
(i) \(a_n = \dfrac{5n}{n + 2}\); find \(a_6\) and \(a_{13}\)
For \(a_6\): Substitute \(n = 6\)
\[a_6 = \dfrac{5(6)}{6 + 2} = \dfrac{30}{8} = \dfrac{15}{4}\]
For \(a_{13}\): Substitute \(n = 13\)
\[a_{13} = \dfrac{5(13)}{13 + 2} = \dfrac{65}{15} = \dfrac{13}{3}\]
\(a_6 = \dfrac{15}{4}\), \(a_{13} = \dfrac{13}{3}\)
(ii) \(a_n = -(n^2 - 4)\); find \(a_4\) and \(a_{11}\)
For \(a_4\): Substitute \(n = 4\)
\[a_4 = -(4^2 - 4) = -(16 - 4) = -12\]
For \(a_{11}\): Substitute \(n = 11\)
\[a_{11} = -(11^2 - 4) = -(121 - 4) = -117\]
\(a_4 = -12\), \(a_{11} = -117\)
5. Find \(a_8\) and \(a_{15}\) whose \(n^{\text{th}}\) term is \(a_n = \begin{cases} \dfrac{n^2 - 1}{n + 3} & ; \text{ if } n \text{ is even}, n \in \mathbb{N} \\ \dfrac{n^2}{2n + 1} & ; \text{ if } n \text{ is odd}, n \in \mathbb{N} \end{cases}\)
To find \(a_8\): Since \(n = 8\) is an even number, use \(a_n = \dfrac{n^2 - 1}{n + 3}\):
\[a_8 = \dfrac{8^2 - 1}{8 + 3} = \dfrac{64 - 1}{11} = \dfrac{63}{11}\]
To find \(a_{15}\): Since \(n = 15\) is an odd number, use \(a_n = \dfrac{n^2}{2n + 1}\):
\[a_{15} = \dfrac{15^2}{2(15) + 1} = \dfrac{225}{30 + 1} = \dfrac{225}{31}\]
\(a_8 = \dfrac{63}{11}\), \(a_{15} = \dfrac{225}{31}\)
6. If \(a_1 = 1\), \(a_2 = 1\) and \(a_n = 2a_{n-1} + a_{n-2}\), \(n \ge 3, n \in \mathbb{N}\), then find the first six terms of the sequence.
Given:
- \(a_1 = 1\)
- \(a_2 = 1\)
Calculate remaining terms using \(a_n = 2a_{n-1} + a_{n-2}\):
- For \(n = 3\): \(a_3 = 2a_2 + a_1 = 2(1) + 1 = 2 + 1 = 3\)
- For \(n = 4\): \(a_4 = 2a_3 + a_2 = 2(3) + 1 = 6 + 1 = 7\)
- For \(n = 5\): \(a_5 = 2a_4 + a_3 = 2(7) + 3 = 14 + 3 = 17\)
- For \(n = 6\): \(a_6 = 2a_5 + a_4 = 2(17) + 7 = 34 + 7 = 41\)
First six terms: 1, 1, 3, 7, 17, 41
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