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If S = 2001/(2*3) + 2001/(3*4) + 2001/(4*5) + ... + 2001/(19*20), what

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Bunuel
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If S = 2001/(2*3) + 2001/(3*4) + 2001/(4*5) + ... + 2001/(19*20), what [#permalink] New post   11 Nov 2022, 09:53
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If \(S = \frac{2001}{2*3} + \frac{2001}{3*4} + \frac{2001}{4*5} + ... + \frac{2001}{19*20}\), what is the value of S ?

A. 700.45
B. 800.55
C. 900.45
D. 1000.75
E. 1100.45
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Kinshook
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If S = 2001/(2*3) + 2001/(3*4) + 2001/(4*5) + ... + 2001/(19*20), what [#permalink] New post   12 Nov 2022, 06:47
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Asked: If \(S = \frac{2001}{2*3} + \frac{2001}{3*4} + \frac{2001}{4*5} + ... + \frac{2001}{19*20}\), what is the value of S ?

\(S = \frac{2001}{2*3} + \frac{2001}{3*4} + \frac{2001}{4*5} + ... + \frac{2001}{19*20}\)

\(t_n = \frac{2001}{(n+1)(n+2)} = \frac{2001{(n+2)-(n+1)}}{(n+2)(n+1)} = 2001 ( \frac{1}{n+1} - \frac{1}{n+2})\)

\(S = 2001 * ( \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + .... \frac{1}{19} - \frac{1}{20}) = 2001 (\frac{1}{2} - \frac{1}{20}) = 2001 *\frac{9}{20} = \frac{18009}{20} = 900.45\\
\)

IMO C
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Re: If S = 2001/(2*3) + 2001/(3*4) + 2001/(4*5) + ... + 2001/(19*20), what [#permalink] New post   12 Feb 2024, 01:31
Bunuel can you pls explain the solution
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Re: If S = 2001/(2*3) + 2001/(3*4) + 2001/(4*5) + ... + 2001/(19*20), what [#permalink] New post   12 Feb 2024, 02:11
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alphaNew13 wrote:
Bunuel can you pls explain the solution

­
Bunuel wrote:
If \(S = \frac{2001}{2*3} + \frac{2001}{3*4} + \frac{2001}{4*5} + ... + \frac{2001}{19*20}\), what is the value of S ?

A. 700.45
B. 800.55
C. 900.45
D. 1000.75
E. 1100.45

­
The point is:

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

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

    \(= \frac{1}{n(n + 1)}\)­
­­

Hence, \(\frac{1}{2*3}\)­ can be written as \(\frac{1}{2} - \frac{1}{3}\)­­­. Similarly, \(\frac{1}{3*4}\) can be written as \(\frac{1}{3} - \frac{1}{4}\). And so on. 

Therefore: 

    \(\frac{2001}{2*3} + \frac{2001}{3*4} + \frac{2001}{4*5} + ... + \frac{2001}{19*20} = \)

    \(=2001(\frac{1}{2*3} + \frac{1}{3*4} + \frac{1}{4*5} + ... + \frac{1}{19*20}) = \)

    \(=2001((\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + (\frac{1}{4} - \frac{1}{5}) + ... + (\frac{1}{19} - \frac{1}{20}))\)­
 ­
Observe that the fractions starting from the second one will be canceled with the following fractions. Thus, we'd be left only with the first fraction, \(\frac{1}{2}\), and with the last fraction, \(- \frac{1}{20}\), because there is no next fraction for it to be canceled by. So, we'd be left:

    \(=2001(\frac{1}{2} - \frac{1}{20})=\)

    \(=2001(\frac{18}{40})=\)

    \(= 900.45\)
    ­

Answer: C.

Hope it helps.


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Re: If S = 2001/(2*3) + 2001/(3*4) + 2001/(4*5) + ... + 2001/(19*20), what [#permalink]
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