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

##### 1 Answer

Jan 7, 2017

I got

It may be easier to convert to improper fractions.

#=> 1, 3/2, 7/4, 15/8, . . . #

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

Notice how if we write a series of representations of

#2/1, 4/2, 8/4, 16/8, 32/16, . . . #

Do you see how the above terms simply subtract

#=> color(blue)(a_n = sum_(n=0)^(N) 2 - 1/(2^n))#

Let's test it!

#=> (2 - 1/(2^0)), (2 - 1/(2^1)), (2 - 1/(2^2)), (2 - 1/(2^3)), . . . #

#=> 1, 3/2, 7/4, 15/8, 31/16, 63/32, . . . # #color(blue)(sqrt"")#