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

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Joined: 02 Sep 2009
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In the sequence a1, a2, a3, ..., a100 the kth term is defined by ak  [#permalink]

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09 May 2018, 01:09
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60% (02:01) correct 40% (02:16) wrong based on 37 sessions

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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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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

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

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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

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

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