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In the sequence a1, a2, a3,....a100, the kth term is defined
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24 Feb 2016, 15:57
24 Feb 2016, 15:57
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Question Stats:
78% (02:36) correct
21% (02:19) wrong based on 95 sessions
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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. \(1\)
A. \(\frac{1}{10100}\)
B. \(\frac{1}{100}\)
C. \(\frac{1}{101}\)
D. \(\frac{100}{101}\)
E. \(1\)
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Joined: 16 May 2014
Posts: 592
Given Kudos: 0
GRE 1: Q165 V161
Re: In the sequence a1, a2, a3,....a100, the kth term is defined
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25 Feb 2016, 04:24
25 Feb 2016, 04:24
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Answer :
D \(\frac{100}{101}\)
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- \(\frac{1}{101}\)
= \(\frac{100}{101}\)
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- \(\frac{1}{101}\)
= \(\frac{100}{101}\)
General Discussion
Re: In the sequence a1, a2, a3,....a100, the kth term is defined
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29 Jan 2019, 00:27
29 Jan 2019, 00:27
3
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?
