Write the first five terms of the sequence whose nth term is aₙ = n(n²+5)/(4).

Exercise 8.1 · Q6

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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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Substituting $n = 1, 2, 3, 4, 5$ into $a_n = n\dfrac{n^2+5}{4}$ gives the first five terms $\dfrac{3}{2}, \dfrac{9}{2}, \dfrac{21}{2}, 21, \dfrac{75}{2}$ .

A sequence given by an explicit formula lets us compute any term directly by substituting its position number $n$ . Here the rule is $a_n = n\dfrac{n^2+5}{4}$ , which is easier to work with once rewritten as a single fraction:

a_n = \dfrac{n(n^2+5)}{4} = \dfrac{n^3+5n}{4}

Term 1 ( $n=1$ ):

a_1 = \dfrac{1^3+5(1)}{4} = \dfrac{1+5}{4} = \dfrac{6}{4} = \dfrac{3}{2}

Term 2 ( $n=2$ ):

a_2 = \dfrac{2^3+5(2)}{4} = \dfrac{8+10}{4} = \dfrac{18}{4} = \dfrac{9}{2}

Term 3 ( $n=3$ ):

a_3 = \dfrac{3^3+5(3)}{4} = \dfrac{27+15}{4} = \dfrac{42}{4} = \dfrac{21}{2}

Term 4 ( $n=4$ ):

a_4 = \dfrac{4^3+5(4)}{4} = \dfrac{64+20}{4} = \dfrac{84}{4} = 21

Term 5 ( $n=5$ ): …

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Q.Write the first five terms of the sequence whose nnnth term is an=n n2+54a_n = n\,\dfrac{n^2+5}{4}an​=n4n2+5​.

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Q.Write the first five terms of the sequence whose $n$ th term is $a_n = n\,\dfrac{n^2+5}{4}$ .

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