# How do you find the nth term of the sequence 1, 1 1/2, 1 3/4, 1 7/8, ...?

Jan 7, 2017

I got ${a}_{n} = {\sum}_{n = 0}^{N} 2 - \frac{1}{{2}^{n}}$.

It may be easier to convert to improper fractions.

$\implies 1 , \frac{3}{2} , \frac{7}{4} , \frac{15}{8} , . . .$

We may recognize that each of these terms is a bit less than $2$.

Notice how if we write a series of representations of $2$ like this, even though they're all $2$, we can see a pattern:

$\frac{2}{1} , \frac{4}{2} , \frac{8}{4} , \frac{16}{8} , \frac{32}{16} , . . .$

Do you see how the above terms simply subtract $\frac{1}{{2}^{n}}$ from $2$? i.e. $\frac{2}{1} - \frac{1}{{2}^{0}} = 1$, $\frac{4}{2} - \frac{1}{{2}^{1}} = \frac{3}{2}$, etc. Therefore:

$\implies \textcolor{b l u e}{{a}_{n} = {\sum}_{n = 0}^{N} 2 - \frac{1}{{2}^{n}}}$

Let's test it!

$\implies \left(2 - \frac{1}{{2}^{0}}\right) , \left(2 - \frac{1}{{2}^{1}}\right) , \left(2 - \frac{1}{{2}^{2}}\right) , \left(2 - \frac{1}{{2}^{3}}\right) , . . .$

$\implies 1 , \frac{3}{2} , \frac{7}{4} , \frac{15}{8} , \frac{31}{16} , \frac{63}{32} , . . .$ color(blue)(sqrt"")