Write the first five terms of the following sequence and obtain the corresponding series: a₁ = 3, aₙ = 3an-1 +

Exercise 8.1 · Q11

a_1 = 3

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Concept understanding — Sequence Terms Evaluation

Sequence Terms Evaluation

A parking garage numbers its levels 1, 2, 3, ... and posts a rule on the wall: "Level $n$ sits $n^2$ metres below street level." You don't need to walk down to level 10 to find its depth -- you simply compute $10^2 = 100$ metres. That one substitution is the entire idea behind sequence terms evaluation: once a sequence has a rule connecting position to value, any term can be found directly, without listing every term before it.

What "Evaluating a Term" Means

A sequence assigns a number to each position $1, 2, 3, \dots$ . The term at position $n$ is written $a_n$ (read "a sub n"). When the sequence is given by an explicit formula $a_n = f(n)$ , evaluating the $k$ -th term simply means substituting $n = k$ into $f$ and simplifying.

Note

This is different from a recursive definition, where $a_n$ is defined using $a_{n-1}$ -- for example, $a_1 = 1,\ a_n = a_{n-1} + 2$ . A recursive rule forces you to compute every earlier term first; an explicit formula lets you jump straight to any term you want.

The Precise Statement

If $\{a_n\}$ is a sequence with $a_n = f(n)$ , then for any positive integer $k$ , the $k$ -th term is

a_k = f(k)

The entire procedure is: replace $n$ with the required index, then simplify the resulting number. No equation-solving, no guesswork -- just substitution.

Common Forms You'll Meet in Exams

Type	Formula	5th term
Linear	$a_n = 3n - 2$	$3(5)-2 = 13$
Quadratic	$a_n = n^2 + 1$	$5^2+1 = 26$
Exponential	$a_n = 2^n$	$2^5 = 32$
Rational	$a_n = \dfrac{n}{n+1}$	$\dfrac{5}{6}$
Alternating	$a_n = (-1)^n$	$(-1)^5 = -1$

Worked Example

Find the 7th term of the sequence $a_n = \dfrac{3n-1}{n^2+2}$ .

The index we need is $k = 7$ .
Substitute $n = 7$ :

a_7 = \frac{3(7)-1}{7^2+2} = \frac{21-1}{49+2} = \frac{20}{51}

Check whether it simplifies -- $\gcd(20, 51) = 1$ , so $\dfrac{20}{51}$ is already in lowest terms and is the final answer. …

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Building the sequence term by term from $a_1=3$ and $a_n=3a_{n-1}+2$ gives the first five terms $3, 11, 35, 107, 323$ , with corresponding series $3+11+35+107+323+\dots$

A recursive rule tells you how to get the next term from the previous one, so the terms must be built up in order, you cannot jump straight to $a_5$ without first knowing $a_4$ .

Term 1: given directly, $a_1 = 3$ .

Term 2: $a_2 = 3a_1+2 = 3(3)+2 = 9+2 = 11$

Term 3: $a_3 = 3a_2+2 = 3(11)+2 = 33+2 = 35$

Term 4: $a_4 = 3a_3+2 = 3(35)+2 = 105+2 = 107$

Term 5: $a_5 = 3a_4+2 = 3(107)+2 = 321+2 = 323$

Watch out

A common mistake is dropping the "+2" and just tripling each term, which would wrongly give $3, 9, 27, 81, 243$ . The "+2" must be added at every single step, not just the first.

The corresponding series is the sum of these terms written with plus signs:

$3 + 11 + 35 + 107 + 323 + \dots$ …

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Q.Write the first five terms of the following sequence and obtain the corresponding series: a1=3a_1 = 3a1​=3, an=3an−1+2a_n = 3a_{n-1} + 2an​=3an−1​+2 for all n>1n > 1n>1.

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Q.Write the first five terms of the following sequence and obtain the corresponding series: $a_1 = 3$ , $a_n = 3a_{n-1} + 2$ for all $n > 1$ .

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