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07 Apr 2023, 08:45
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E
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Difficulty:
65%
(hard)
Question Stats:
61% (02:37) correct
39%
(02:51)
wrong
based on 114
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If \((n+1)!*f(n) = (n-1)!\), what is the value of\(f(1) + f(2) + … + f(100)\)?
A. \(\frac{1}{99}\)
B. \(\frac{1}{100}\)
C. \(\frac{1}{101}\)
D. \(\frac{99}{100}\)
E. \(\frac{100}{101}\)
A. \(\frac{1}{99}\)
B. \(\frac{1}{100}\)
C. \(\frac{1}{101}\)
D. \(\frac{99}{100}\)
E. \(\frac{100}{101}\)
07 Apr 2023, 13:38
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If \((n+1)!*f(n) = (n-1)!\), what is the value of\(f(1) + f(2) + … + f(100)\)?
A. \(\frac{1}{99}\)
B. \(\frac{1}{100}\)
C. \(\frac{1}{101}\)
D. \(\frac{99}{100}\)
E. \(\frac{100}{101}\)
A. \(\frac{1}{99}\)
B. \(\frac{1}{100}\)
C. \(\frac{1}{101}\)
D. \(\frac{99}{100}\)
E. \(\frac{100}{101}\)
\((n+1)!*f(n) = (n-1)!\) tells us that \(f(n) = \frac{(n-1)!}{(n+1)!}\)
\(f(1)=\frac{(n-1)!}{(n+1)!}=\frac{0!}{2!}=\frac{1}{2}\) To that, we are going to add more positive numbers, so we can eliminate answer choices A, B, and C.
The terms will be:
\(\frac{(n-1)!}{(n+1)!}+\frac{(n)!}{(n+2)!}+\frac{(n+1)!}{(n+3)!}+\frac{(n+2)!}{(n+4)!}. . . \)
The denominator of the 100th term will be (n+100)!, so 101!. The 101 will need to stay in the denominator since it is prime and there is no other 101 that will appear in the numerator to offset it. Only one remaining answer choice has 101 in the denominator.
Answer choice E.
26 Nov 2023, 09:11
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(n+1)!*f(n) = (n-1)! and we need to find the value of\(f(1) + f(2) + … + f(100)\)?
(n+1)!*f(n) = (n-1)!
=> \(f(n) = \frac{(n-1)!}{(n+1)!}\) = \(\frac{(n-1)!}{(n+1)*n*(n-1)!}\) = \(\frac{1}{n*(n+1)}\)
= \(\frac{n+1 - n }{ n*(n+1)}\) = \(\frac{1}{n - 1/n+1}\)
=> f(1) + f(2) + … + f(100) = \(\frac{1}{1} - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + ... + \frac{1}{100} - \frac{1}{101}\)
Alternate terms will cancel out and we will be left with only the first and the last terms
= \(1 - \frac{1}{101}\) = \(\frac{101 - 1}{101}\) = \(\frac{100}{101}\)
So, Answer will be E
Hope it helps!
Watch the following video to learn the Basics of Functions and Custom Characters
(n+1)!*f(n) = (n-1)!
=> \(f(n) = \frac{(n-1)!}{(n+1)!}\) = \(\frac{(n-1)!}{(n+1)*n*(n-1)!}\) = \(\frac{1}{n*(n+1)}\)
= \(\frac{n+1 - n }{ n*(n+1)}\) = \(\frac{1}{n - 1/n+1}\)
=> f(1) + f(2) + … + f(100) = \(\frac{1}{1} - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + ... + \frac{1}{100} - \frac{1}{101}\)
Alternate terms will cancel out and we will be left with only the first and the last terms
= \(1 - \frac{1}{101}\) = \(\frac{101 - 1}{101}\) = \(\frac{100}{101}\)
So, Answer will be E
Hope it helps!
Watch the following video to learn the Basics of Functions and Custom Characters
