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In the sequence a1, a2, a3,....a100, the kth term is defined

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Intern
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In the sequence a1, a2, a3,....a100, the kth term is defined [#permalink] New post 24 Feb 2016, 15:57
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In the sequence a1, a2, a3,....a100, the kth term is defined as \(ak= \frac{1}{k} - \frac{1}{k+1}\) for all integers k from 1 through 100. What is the sum of 100 terms of the sequence?

A. \(\frac{1}{10100}\)

B. \(\frac{1}{100}\)

C. \(\frac{1}{101}\)

D. \(\frac{100}{101}\)

E. \(1\)
[Reveal] Spoiler: OA
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Re: In the sequence a1, a2, a3,....a100, the kth term is defined [#permalink] New post 25 Feb 2016, 04:24
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Answer :
D \(\frac{100}{101}\)

Explanation:


Given
a_k = \(\frac{1}{k}\)-\(\frac{1}{k+1}\);

So the sum of a_1 to a_100 is explained below-
a_1 + a_2 + a_3 + ...... + a_100
= \(\frac{1}{1}\) - \(\frac{1}{2}\) + \(\frac{1}{2}\) - \(\frac{1}{3}\)+ ..... + \(\frac{1}{100}\) - \(\frac{1}{101}\)

= 1- \(\frac{1}{101}\)

= \(\frac{100}{101}\)
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Re: In the sequence a1, a2, a3,....a100, the kth term is defined [#permalink] New post 29 Jan 2019, 00:27
hello, I do not understand from the point of 1-1/101. How do you come to a conclusion that the sum is 1-1/101?
Re: In the sequence a1, a2, a3,....a100, the kth term is defined   [#permalink] 29 Jan 2019, 00:27
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In the sequence a1, a2, a3,....a100, the kth term is defined

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