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16 Sep 2014, 01:22
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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. \(\frac{T + 1}{100}\)
A. \(\frac{T}{101}\)
B. \(\frac{T}{100} + 100\)
C. \(\frac{100 + T}{T}\)
D. \(\frac{102}{T}\)
E. \(\frac{T + 1}{100}\)
16 Sep 2014, 01:22
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Official Solution:
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
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
19 Oct 2023, 05:11
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
