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27 Feb 2013, 13:39
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D
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80% (01:47) correct
20%
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If \(S\) is the product of the integers from 1 to 100, inclusive, and \(T\) is the product of the integers from 1 to 101, inclusive, what is \(\frac{1}{S} + \frac{1}{T}\) in terms of \(T\) ?
A. \(\frac{T}{101}\)
B. \(\frac{T}{100} + 100\)
C. \(\frac{100 + T}{T}\)
D. \(\frac{102}{T}\)
E. \(\frac{T + 1}{100}\)
M24-35
A. \(\frac{T}{101}\)
B. \(\frac{T}{100} + 100\)
C. \(\frac{100 + T}{T}\)
D. \(\frac{102}{T}\)
E. \(\frac{T + 1}{100}\)
M24-35
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27 Feb 2013, 14:54
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If \(S\) is the product of the integers from 1 to 100, inclusive, and \(T\) is the product of the integers from 1 to 101, inclusive, what is \(\frac{1}{S} + \frac{1}{T}\) in terms of \(T\) ?
A. \(\frac{T}{101}\)
B. \(\frac{T}{100} + 100\)
C. \(\frac{100 + T}{T}\)
D. \(\frac{102}{T}\)
E. \(\frac{T + 1}{100}\)
Given: \(S=100!\) and \(T=101!\).
\(\frac{1}{S}+\frac{1}{T}=\)
\(=\frac{1}{100!}+\frac{1}{101!}=\)
\(=\frac{101}{101!}+\frac{1}{101!}=\)
\(=\frac{101+1}{101!}=\)
\(=\frac{102}{T}\).
Answer: D
A. \(\frac{T}{101}\)
B. \(\frac{T}{100} + 100\)
C. \(\frac{100 + T}{T}\)
D. \(\frac{102}{T}\)
E. \(\frac{T + 1}{100}\)
Given: \(S=100!\) and \(T=101!\).
\(\frac{1}{S}+\frac{1}{T}=\)
\(=\frac{1}{100!}+\frac{1}{101!}=\)
\(=\frac{101}{101!}+\frac{1}{101!}=\)
\(=\frac{101+1}{101!}=\)
\(=\frac{102}{T}\).
Answer: D
General Discussion
27 Feb 2013, 20:56
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aaroncorr
You can also substitute T for S right away to get the expression in terms of T.
Given: S=100! and T=101! Therefore, S = T/101
1/S+1/T = 101/T + 1/T = 102/T
