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01 Nov 2023, 02:10
Kudos
Bookmarks
D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
50% (02:38) correct
50%
(02:21)
wrong
based on 117
sessions
History
Date
Time
Result
Not Attempted Yet
What is the value of the sum \(\frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!} + · · · + \frac{99}{100!}\)?
(A) \(\frac{1}{2} − \frac{1}{99!}\)
(B) \(\frac{1}{2} − \frac{1}{100!}\)
(C) \(1 − \frac{1}{99!}\)
(D) \(1 − \frac{1}{100!}\)
(E) \(1\)
(A) \(\frac{1}{2} − \frac{1}{99!}\)
(B) \(\frac{1}{2} − \frac{1}{100!}\)
(C) \(1 − \frac{1}{99!}\)
(D) \(1 − \frac{1}{100!}\)
(E) \(1\)
01 Nov 2023, 08:35
Kudos
Bookmarks
Bunuel
OA is D
Posted from my mobile device
Attachments
1000057706.jpg [ 1.41 MiB | Viewed 2969 times ]
General Discussion
01 Nov 2023, 02:21
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an = n/(n+1)!
= ((n+1)-1)/(n+1)!= 1/n! - 1/(n+1)!
Sn= 1/1!-1/2! + 1/2! - 1/3! +......+1/99! - 1/100!
So finally we will be left with
1- 1/100!
Posted from my mobile device
= ((n+1)-1)/(n+1)!= 1/n! - 1/(n+1)!
Sn= 1/1!-1/2! + 1/2! - 1/3! +......+1/99! - 1/100!
So finally we will be left with
1- 1/100!
Posted from my mobile device
