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

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In the sequence a1, a2, a3, ..., a100 the kth term is defined by ak  [#permalink]

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New post 09 May 2018, 01:09
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In the sequence a1, a2, a3, ..., a100 the kth term is defined by \(a_k = \frac{1}{k} - \frac{1}{k+1}\) for all integers k from 1 through 100. What is the sum of the 100 terms of this sequence?

A) \(\frac{1}{10,100}\)

B) \(\frac{1}{101}\)

C) \(\frac{1}{100}\)

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

E) \(1\)

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Re: In the sequence a1, a2, a3, ..., a100 the kth term is defined by ak  [#permalink]

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New post 09 May 2018, 04:19
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It is given that ak=1/k−1/k+1

a1= 1/1-1/2

a2= 1/2-1/3

a3= 1/3-1/4
....
..
.
a99=1/99-1/100

a100= 1/100-1/101

When we add all the terms, it can be seen that all the terms from 1/2 to 1/100 will cancel each other because of opposite signs. Only remaining terms will be 1/1 and -1/101.

So sum = 1/1-1/101= 101-1/101= 100/101

Answer is D
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Re: In the sequence a1, a2, a3, ..., a100 the kth term is defined by ak  [#permalink]

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New post 10 May 2018, 10:37
Bunuel wrote:
In the sequence a1, a2, a3, ..., a100 the kth term is defined by \(a_k = \frac{1}{k} - \frac{1}{k+1}\) for all integers k from 1 through 100. What is the sum of the 100 terms of this sequence?

A) \(\frac{1}{10,100}\)

B) \(\frac{1}{101}\)

C) \(\frac{1}{100}\)

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

E) \(1\)


Starting with the first 4 terms we have:

a1 + a2 + a3 + a4 = (1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + (1/4 - 1/5)

We see that all terms cancel except for the first term and the final term. Combining those two terms, we have 1 - 1/5 = 4/5.

Extending this pattern to all 100 numbers, we have:

1 - 1/101 = 101/101 - 1/101 = 100/101

Answer: D
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Re: In the sequence a1, a2, a3,....a100, the kth term is defined as [m]ak=  [#permalink]

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New post 31 Jan 2019, 23:00
Bunuel wrote:
In the sequence a1, a2, a3,....a100, the kth term is defined as \(a_k= \frac{1}{k} - \frac{1}{k+1}\) 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. \(1\)


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

\(a_1= \frac{1}{1} - \frac{1}{1+1}\)

\(a_2= \frac{1}{2} - \frac{1}{2+1}\)

\(a_100= \frac{1}{100} - \frac{1}{100+1}\)

Sum = (1/1) - (1/2) + (1/2) - (1/3) ...... (1/100+1/101 )

Sum = 1- (1/101)

Sum = 100/101

Hence D
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Re: In the sequence a1, a2, a3,....a100, the kth term is defined as [m]ak=   [#permalink] 31 Jan 2019, 23:00
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