How do you write the nth term rule for the sequence 1, 3, 5, 7, 9, ...?

Sep 11, 2016

${T}_{n} = 2 n - 1$

Explanation:

This is clearly an arithmetic sequence because the terms differ by 2 each time.

To find the nth term rule we need:
a value for the *first term , a
and a value for the *common difference d
.

These values are then plugged into the formula:

${T}_{n} = a + \left(n - 1\right) d$

$a = 1 \mathmr{and} d = 2$

${T}_{n} = a + \left(n - 1\right) d$

${T}_{n} = 1 + \left(n - 1\right) 2$

${T}_{n} = 1 + 2 n - 2$

${T}_{n} = 2 n - 1$

Sep 11, 2016

${a}_{n} = \left(2 n - 1\right)$ for $n \in \mathbb{N}$
This is the sequence of the odd positive integers where the ${n}^{t h}$ term $\left({a}_{n}\right)$ is:
${a}_{n} = \left(2 n - 1\right)$ for $n \in \mathbb{N}$