Write the first five terms of each of the sequences in Exercises 1 to 6 whose nth terms are:
Q1. \(a_n = n (n + 2)\)
Solution
The $n$th term of the sequence is given by \(a_n = n(n + 2)\). To find the first five terms, we substitute \(n = 1, 2, 3, 4, 5\) successively into this formula. Each computation multiplies $n$ by the next odd number after $2n - 1$, generating a quadratic sequence.
$$ \begin{aligned} a_{n} &= n(n + 2) \\ a_{1} &= 1(1 + 2) = 1 \times 3 = 3 \\ a_{2} &= 2(2 + 2) = 2 \times 4 = 8 \\ a_{3} &= 3(3 + 2) = 3 \times 5 = 15 \\ a_{4} &= 4(4 + 2) = 4 \times 6 = 24 \\ a_{5} &= 5(5 + 2) = 5 \times 7 = 35 \end{aligned} $$
Thus, the first five terms of the sequence are $3, 8, 15, 24, 35$. These values confirm the pattern of the quadratic expression, where each term increases by successive even numbers starting from $5$.
Q2. \(a_n=\dfrac{n}{n+1}\)
Solution
The \(n\)th term of the sequence is given by \(\ a_n = \dfrac{n}{n+1}\). To find the first five terms, we substitute \(n = 1, 2, 3, 4, 5\) into this rational function. Each term represents the ratio of consecutive integers, forming a sequence that approaches 1 asymptotically.
$$ \begin{aligned} a_n &= \dfrac{n}{n+1} \\ a_1 &= \dfrac{1}{1+1} = \dfrac{1}{2} \\ a_2 &= \dfrac{2}{2+1} = \dfrac{2}{3} \\ a_3 &= \dfrac{3}{3+1} = \dfrac{3}{4} \\ a_4 &= \dfrac{4}{4+1} = \dfrac{4}{5} \\ a_5 &= \dfrac{5}{5+1} = \dfrac{5}{6} \end{aligned} $$
Thus, the first five terms are \(\ \dfrac{1}{2}, \dfrac{2}{3}, \dfrac{3}{4}, \dfrac{4}{5}, \dfrac{5}{6}\). This sequence can also be expressed as \(\ a_n = 1 - \dfrac{1}{n+1}\), revealing its harmonic nature where each term gets closer to 1.
Q3. \(a_n=2^n\)
Solution
The \(n\)th term of the sequence is given by \(a_n = 2^n\). This is a geometric sequence where each term is obtained by doubling the previous term. To find the first five terms, compute \(2^n\) for \(n = 1\) through \(n = 5\).
$$ \begin{aligned} a_n &= 2^n \\ a_1 &= 2^1 = 2 \\ a_2 &= 2^2 = 4 \\ a_3 &= 2^3 = 8 \\ a_4 &= 2^4 = 16 \\ a_5 &= 2^5 = 32 \end{aligned} $$
Thus the first five terms are \(2, 4, 8, 16, 32\). Each term doubles the preceding one, characteristic of exponential growth with common ratio \(2\).
Q4. \(a_n=\dfrac{2n-3}{6}\)
Solution
The \(n\)th term of the sequence is given by \(a_n = \dfrac{2n-3}{6}\). This is an arithmetic sequence in disguise since it's linear in \(n\). To find the first five terms, substitute \(n = 1, 2, 3, 4, 5\) into the formula successively.
$$ \begin{aligned} a_n &= \dfrac{2n-3}{6} \\ a_1 &= \dfrac{2\cdot1-3}{6} = \dfrac{-1}{6} \\ a_2 &= \dfrac{2\cdot2-3}{6} = \dfrac{1}{6} \\ a_3 &= \dfrac{2\cdot3-3}{6} = \dfrac{3}{6} = \dfrac{1}{2} \\ a_4 &= \dfrac{2\cdot4-3}{6} = \dfrac{5}{6} \\ a_5 &= \dfrac{2\cdot5-3}{6} = \dfrac{7}{6} \end{aligned} $$
Thus the first five terms are \(-\dfrac{1}{6}, \dfrac{1}{6}, \dfrac{1}{2}, \dfrac{5}{6}, \dfrac{7}{6}\). Notice the common difference of \(\dfrac{1}{3}\) between consecutive terms, confirming the arithmetic nature of this linear sequence.
Q5. \( a_n = (–1)^{n–1}\; 5^{n+1}\)
Solution
The \(n\)th term of the sequence is given by \(a_n = (-1)^{n-1} \cdot 5^{n+1}\). This alternating geometric sequence multiplies powers of 5 by \((-1)^{n-1}\), which is \(+1\) for odd \(n\) and \(-1\) for even \(n\). Compute the first five terms by direct substitution.
$$ \begin{aligned} a_n &= (-1)^{n-1} \cdot 5^{n+1} \\ a_1 &= (-1)^{0} \cdot 5^{2} = 25 \\ a_2 &= (-1)^{1} \cdot 5^{3} = -125 \\ a_3 &= (-1)^{2} \cdot 5^{4} = 625 \\ a_4 &= (-1)^{3} \cdot 5^{5} = -3125 \\ a_5 &= (-1)^{4} \cdot 5^{6} = 15625 \end{aligned} $$
Thus the first five terms are \(25, -125, 625, -3125, 15625\). The signs alternate starting positive, while magnitudes grow exponentially by factor of \(5^2 = 25\) each time due to the combined power increase.
Q6. \(a_n=n\dfrac{n^2+5}{4}\)
Solution
The \(n\)th term of the sequence is given by \(a_n = n\dfrac{n^2 + 5}{4}\). This cubic expression generates terms that are always half-integers or integers. To find the first five terms, substitute \(n = 1, 2, 3, 4, 5\) step by step into the formula.
$$ \begin{aligned} a_n &= n\dfrac{n^2 + 5}{4} \\ a_1 &= 1\cdot\dfrac{1^2 + 5}{4} = \dfrac{6}{4} = \dfrac{3}{2} \\ a_2 &= 2\cdot\dfrac{2^2 + 5}{4} = 2\cdot\dfrac{9}{4} = \dfrac{9}{2} \\ a_3 &= 3\cdot\dfrac{3^2 + 5}{4} = 3\cdot\dfrac{14}{4} = \dfrac{21}{2} \\ a_4 &= 4\cdot\dfrac{4^2 + 5}{4} = 4\cdot\dfrac{21}{4} = 21 \\ a_5 &= 5\cdot\dfrac{5^2 + 5}{4} = 5\cdot\dfrac{30}{4} = \dfrac{75}{2} \end{aligned} $$
Thus the first five terms are \(\dfrac{3}{2}, \dfrac{9}{2}, \dfrac{21}{2}, 21, \dfrac{75}{2}\). Notice the pattern where most terms have denominator 2, except \(a_4\) which simplifies perfectly to an integer.
Find the indicated terms in each of the sequences in Exercises 7 to 10 whose nth terms are:
Q7. \(a_n = 4n – 3;\; a_{17},\; a_{24}\)
Solution
The \(n\)th term of the arithmetic sequence is given by \(a_n = 4n - 3\). This linear formula has first term \(a_1 = 1\) and common difference \(d = 4\). To find \(a_{17}\) and \(a_{24}\), substitute the given indices directly.
$$ \begin{aligned} a_n &= 4n - 3 \\ a_{17} &= 4 \times 17 - 3 = 68 - 3 = 65 \\ a_{24} &= 4 \times 24 - 3 = 96 - 3 = 93 \end{aligned} $$
Thus \(a_{17} = 65\) and \(a_{24} = 93\). These terms follow the pattern where each subsequent term increases by 4, confirming the arithmetic progression.
Q8. \(a_n=\dfrac{n^2}{2^n};\;a_{7}\)
Solution
The \(n\)th term of the sequence is given by \(a_n = \dfrac{n^2}{2^n}\). This converges to zero as \(n\) grows since the exponential denominator outpaces the quadratic numerator. To find \(a_7\), substitute \(n = 7\) directly into the formula.
$$ \begin{aligned} a_n &= \dfrac{n^2}{2^n} \\ a_7 &= \dfrac{7^2}{2^7} = \dfrac{49}{128} \end{aligned} $$
Thus \(a_7 = \dfrac{49}{128}\). The fraction is already in simplest terms as 49 and 128 share no common factors beyond 1.
Q9. \(a_n=(-1)^{n-1}n^3;\; a_9\)
Solution
The \(n\)th term of the sequence is given by \(a_n = (-1)^{n-1} n^3\). This alternates signs based on whether \(n\) is odd or even, multiplying cubes by \(+1\) for odd \(n\) and \(-1\) for even \(n\). To find \(a_9\), substitute \(n = 9\) since 9 is odd.
$$ \begin{aligned} a_n &= (-1)^{n-1} n^3 \\ a_9 &= (-1)^{8} \cdot 9^3 = 1 \cdot 729 = 729 \end{aligned} $$
Thus \(a_9 = 729\). Since \((-1)^8 = 1\), the term remains positive, matching the pattern for odd indices.
Q10. \(a_n=\dfrac{n(n-2)}{n+3};\;a_{20}\)
Solution
The \(n\)th term of the sequence is given by \(a_n = \dfrac{n(n-2)}{n+3}\). This rational function simplifies to approximately \(n - 2 - \dfrac{1}{n+3}\) for large \(n\), approaching linear growth. To find \(a_{20}\), substitute \(n = 20\) directly.
$$ \begin{aligned} a_n &= \dfrac{n(n-2)}{n+3} \\ a_{20} &= \dfrac{20(20-2)}{20+3} = \dfrac{20 \times 18}{23} = \dfrac{360}{23} \end{aligned} $$
Thus \(a_{20} = \dfrac{360}{23}\). The fraction is in simplest terms as 360 and 23 share no common factors, avoiding unnecessary decimal approximation.
Write the first five terms of each of the sequences in Exercises 11 to 13 and obtain the corresponding series:
Q11. \(a_1 = 3, an = 3a_{n – 1} + 2\text{ for all } n \gt 1\)
Solution
The sequence is defined recursively by \(a_1 = 3\) and \(a_n = 3a_{n-1} + 2\) for \(n > 1\). This linear recurrence generates terms by multiplying the previous term by 3 and adding 2 each time. Compute the first five terms iteratively to identify the pattern.
$$ \begin{aligned} a_1 &= 3 \\ a_2 &= 3a_1 + 2 = 3 \times 3 + 2 = 11 \\ a_3 &= 3a_2 + 2 = 3 \times 11 + 2 = 35 \\ a_4 &= 3a_3 + 2 = 3 \times 35 + 2 = 107 \\ a_5 &= 3a_4 + 2 = 3 \times 107 + 2 = 323 \end{aligned} $$
Thus the first five terms are \(3, 11, 35, 107, 323\). The corresponding series is \(3 + 11 + 35 + 107 + 323\). Each term grows rapidly due to the multiplicative factor of 3 in the recurrence relation.
Q12. \(a_1 = – 1, a_n=\dfrac{a_{n-1}}{n},\;n\geq2\)
Solution
The sequence begins with \(a_1 = -1\) and follows the recursive rule \(a_n = \dfrac{a_{n-1}}{n}\) for \(n \geq 2\). Each term divides the previous by the current index, yielding negative reciprocals of factorials. Compute the first five terms step by step.
$$ \begin{aligned} a_1 &= -1 \\ a_2 &= \dfrac{a_1}{2} = -\dfrac{1}{2} \\ a_3 &= \dfrac{a_2}{3} = -\dfrac{1}{6} \\ a_4 &= \dfrac{a_3}{4} = -\dfrac{1}{24} \\ a_5 &= \dfrac{a_4}{5} = -\dfrac{1}{120} \end{aligned} $$
Thus the first five terms are \(-1, -\dfrac{1}{2}, -\dfrac{1}{6}, -\dfrac{1}{24}, -\dfrac{1}{120}\). The corresponding series is \(-1 - \dfrac{1}{2} - \dfrac{1}{6} - \dfrac{1}{24} - \dfrac{1}{120}\), recognizing \(a_n = -\dfrac{1}{n!}\).
Q13. \(a_1 = a_2 = 2, a_n = a_{n – 1}–1,\; n \gt 2\)
Solution
The sequence starts with \(a_1 = a_2 = 2\) and follows \(a_n = a_{n-1} - 1\) for \(n > 2\). This arithmetic recurrence subtracts 1 from the previous term each time after the second. Compute the first five terms iteratively from the initial values.
$$ \begin{aligned} a_1 &= 2 \\ a_2 &= 2 \\ a_3 &= a_2 - 1 = 1 \\ a_4 &= a_3 - 1 = 0 \\ a_5 &= a_4 - 1 = -1 \end{aligned} $$
Thus the first five terms are \(2, 2, 1, 0, -1\). The corresponding series is \(2 + 2 + 1 + 0 + (-1)\), decreasing by 1 starting from the third term with common difference \(-1\).
Q14. The Fibonacci sequence is defined by \(1 = a_1 = a_2\text{ and }a_n = a_{n – 1} + a_{n – 2} ,\; n \gt 2.\) Find \(\dfrac{a_{n+1}}{a_n}\) for \(n=1,\; 2,\; 3,\; 4,\; 5\)
Solution
The Fibonacci sequence starts with \(a_1 = 1\), \(a_2 = 1\), and \(a_n = a_{n-1} + a_{n-2}\) for \(n > 2\). The task requires computing the ratios \(\dfrac{a_{n+1}}{a_n}\) for \(n = 1\) to \(5\), first generating necessary terms up to \(a_6\). These ratios approach the golden ratio \(\phi \approx 1.618\) as \(n\) increases.
$$ \begin{aligned} a_1 &= 1 \\ a_2 &= 1 \\ a_3 &= a_2 + a_1 = 2 \\ a_4 &= a_3 + a_2 = 3 \\ a_5 &= a_4 + a_3 = 5 \\ a_6 &= a_5 + a_4 = 8 \end{aligned} $$
$$ \begin{aligned} n=1: & \quad \dfrac{a_2}{a_1} = \dfrac{1}{1} = 1 \\ n=2: & \quad \dfrac{a_3}{a_2} = \dfrac{2}{1} = 2 \\ n=3: & \quad \dfrac{a_4}{a_3} = \dfrac{3}{2} \\ n=4: & \quad \dfrac{a_5}{a_4} = \dfrac{5}{3} \\ n=5: & \quad \dfrac{a_6}{a_5} = \dfrac{8}{5} \end{aligned} $$