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

Answer 1
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"")#
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Answer 2

To find the nth term of the sequence, you can observe that each term is obtained by adding 1 to the previous term and then halving the result. The first term is 1. The second term is 1 + 1 = 2, halved to get 1 1/2. The third term is 1 1/2 + 1 = 2 1/2, halved to get 1 3/4. The pattern continues in the same manner. Therefore, the nth term of the sequence can be expressed as ( 1 + \frac{1}{2^n} ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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