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

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Answer 1 · 2017-01-07T02:50:49

I got #a_n = sum_(n=0)^(N) 2 - 1/(2^n)#.

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 #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:

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

Do you see how the above terms simply subtract #1/(2^n)# from #2#? i.e. #2/1 - 1/(2^0) = 1#, #4/2 - 1/(2^1) = 3/2#, etc. Therefore:

#=> 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"")#