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What is the value of 1/(1*2)+1/(2*3)+...+1/(99*100)?

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What is the value of 1/(1*2)+1/(2*3)+...+1/(99*100)?  [#permalink]

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New post Updated on: 28 Nov 2016, 05:08
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What is the value of \(A=\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}+...+\frac{1}{99 \times 100}\)

A. \(\frac{98}{99}\)

B. \(\frac{99}{100}\)

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

D. \(\frac{98}{101}\)

E. \(\frac{99}{101}\)

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Originally posted by broall on 28 Nov 2016, 03:49.
Last edited by broall on 28 Nov 2016, 05:08, edited 1 time in total.
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Re: What is the value of 1/(1*2)+1/(2*3)+...+1/(99*100)?  [#permalink]

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New post 28 Nov 2016, 08:05
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nguyendinhtuong wrote:
What is the value of \(A=\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}+...+\frac{1}{99 \times 100}\)

A. \(\frac{98}{99}\)

B. \(\frac{99}{100}\)

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

D. \(\frac{98}{101}\)

E. \(\frac{99}{101}\)


Let's find the sum of various fractions and look for a pattern

\(\frac{1}{1 \times 2}+\frac{1}{2 \times 3}\) = 1/2 + 1/6
= 3/6 + 1/6
= 4/6
= 2/3


\(\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}\) = 2/3 + \(\frac{1}{3 \times 4}\)
= 2/3 + 1/12
= 8/12 + 1/12
= 9/12
= 3/4


\(\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}+\frac{1}{4 \times 5}\) = 3/4 + \(\frac{1}{4 \times 5}\)
= 3/4 + 1/20
= 15/20 + 1/20
= 16/2
= 4/5

As we can see, the pattern is...
\(A=\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}+...+\frac{1}{(n) \times (n+1)}\) = \(\frac{n}{n+1}\)

So, \(\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}+...+\frac{1}{99 \times 100}\) = \(\frac{99}{100}\)

Answer:

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Re: What is the value of 1/(1*2)+1/(2*3)+...+1/(99*100)?  [#permalink]

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New post 28 Nov 2016, 05:06
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Re: What is the value of 1/(1*2)+1/(2*3)+...+1/(99*100)?  [#permalink]

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New post 28 Nov 2016, 11:27
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Re: What is the value of 1/(1*2)+1/(2*3)+...+1/(99*100)?  [#permalink]

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New post 29 Nov 2016, 15:53
nguyendinhtuong wrote:
What is the value of \(A=\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}+...+\frac{1}{99 \times 100}\)

A. \(\frac{98}{99}\)

B. \(\frac{99}{100}\)

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

D. \(\frac{98}{101}\)

E. \(\frac{99}{101}\)


We begin by noting that 1/(1x2) = 1/1 - 1/2, 1/(2x3) = 1/2 - 1/3, 1/(3x4) = 1/3 - 1/4, … , 1/(99x100) = 1/99 - 1/100.

If we substitute each term in the definition of A by the above, we obtain:

A = 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + … + 1/99 - 1/100

Note that all of the terms (besides 1/1 = 1 and 1/100) cancel; therefore,

A = 1/1 - 1/100 = 99/100

Answer: B
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Re: What is the value of 1/(1*2)+1/(2*3)+...+1/(99*100)?  [#permalink]

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New post 05 Dec 2016, 18:10
Although I can follow the explanations made in this thread I would not be able to come up with them on my own. How can we make sure that we identify the right pattern in such a question?
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Re: What is the value of 1/(1*2)+1/(2*3)+...+1/(99*100)?  [#permalink]

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New post 05 Dec 2016, 18:14
VS93 wrote:
Although I can follow the explanations made in this thread I would not be able to come up with them on my own. How can we make sure that we identify the right pattern in such a question?

In general when I see a question with summation of products (numerator or denominator) I try to see if I can express the general term as a difference. This would help us cancel the terms in a summation series...

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Re: What is the value of 1/(1*2)+1/(2*3)+...+1/(99*100)?  [#permalink]

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New post 15 Dec 2016, 07:18
n=1: 1/2
n=2: 1/2 + 1/2x3 = 1/2 + 1/6 = 2/3
n=3: 4/6 + 1/3x4 = 4/6 + 1/12 = 3/4
n=4: 3/4 + 1/4x5 = 3/4 + 1/20 = 4/5

So, we can deduce that: n(x) = (x-1/x) + (1/x(x+1))

Therefore, n(99) = (98/99) + (1/99*100) = 9801/9900 = 99/100
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Re: What is the value of 1/(1*2)+1/(2*3)+...+1/(99*100)?  [#permalink]

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New post 01 Jan 2017, 09:09
nguyendinhtuong wrote:
What is the value of \(A=\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}+...+\frac{1}{99 \times 100}\)

A. \(\frac{98}{99}\)

B. \(\frac{99}{100}\)

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

D. \(\frac{98}{101}\)

E. \(\frac{99}{101}\)


No need to use pen /pencil
Ex :
1/(1*2) + 1/(2*3)=2/3
Similarly if u expand it you will find 3/4,4/5 going through.
Hence ans last digit 99/100
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Re: What is the value of 1/(1*2)+1/(2*3)+...+1/(99*100)?  [#permalink]

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New post 26 Jul 2017, 02:00
nguyendinhtuong wrote:
What is the value of \(A=\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}+...+\frac{1}{99 \times 100}\)

A. \(\frac{98}{99}\)

B. \(\frac{99}{100}\)

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

D. \(\frac{98}{101}\)

E. \(\frac{99}{101}\)

Best way to solve these kind of questions is to simplify them to a known form.
1/2 + 1/(2x3) + 1/(3x4) + .............+ 1/(99x100)

since 1/(2x3)= 1/6 = 1/2-1/3, In a similar way all terms can be written in to this form
=1/2 + 1/2 -1/3 + 1/3 - 1/4 + 1/4- 1/5+............-1/99+1/99-1/100
=1/2 + 1/2 -1/100 (all others cancelled)
=1-1/100 = 99/100
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Re: What is the value of 1/(1*2)+1/(2*3)+...+1/(99*100)?   [#permalink] 26 Jul 2017, 02:00
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