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

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In the sequence a1, a2, a3,....a100, the kth term is defined [#permalink]  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. $$1$$
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Re: In the sequence a1, a2, a3,....a100, the kth term is defined [#permalink]  25 Feb 2016, 04:24
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Expert's post
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]  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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