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11 Nov 2022, 09:53
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Difficulty:
65%
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Question Stats:
57% (02:11) correct
43%
(02:42)
wrong
based on 74
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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
A. 700.45
B. 800.55
C. 900.45
D. 1000.75
E. 1100.45
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
\(S = \frac{2001}{2*3} + \frac{2001}{3*4} + \frac{2001}{4*5} + ... + \frac{2001}{19*20}\)
\(t_n = \frac{2001}{(n+1)(n+2)} = \frac{2001{(n+2)-(n+1)}}{(n+2)(n+1)} = 2001 ( \frac{1}{n+1} - \frac{1}{n+2})\)
\(S = 2001 * ( \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + .... \frac{1}{19} - \frac{1}{20}) = 2001 (\frac{1}{2} - \frac{1}{20}) = 2001 *\frac{9}{20} = \frac{18009}{20} = 900.45\\
\)
IMO C
alphaNew13
12 Feb 2024, 01:31
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Bunuel can you pls explain the solution
