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18 Nov 2023, 06:50
Kudos
Bookmarks
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
66% (02:39) correct
34%
(02:48)
wrong
based on 269
sessions
History
Date
Time
Result
Not Attempted Yet
What is the value of \((1 - \frac{1}{2^2})(1 - \frac{1}{3^2})(1 - \frac{1}{4^2})….(1 - \frac{1}{20^2})\)?
A. \(\frac{11}{10}\)
B. \(\frac{20}{39} \)
C. \(\frac{21}{40 }\)
D. \(\frac{23}{41} \)
E. \(\frac{27}{50} \)
A. \(\frac{11}{10}\)
B. \(\frac{20}{39} \)
C. \(\frac{21}{40 }\)
D. \(\frac{23}{41} \)
E. \(\frac{27}{50} \)
24 Nov 2023, 01:20
Kudos
Bookmarks
Bunuel
When you look at the first few terms, you see that many factors are getting cancelled.
\(\frac{3}{4} * \frac{8}{9} * \frac{15}{16} * \frac{24}{25}...\)
But what we need is the logic to know what will remain.
\(1 - \frac{1}{n^2} = \frac{(n^2 - 1)}{n^2} = \frac{(n-1)}{n} * \frac{(n+1)}{n}\\
\)
When n = 2, we get \(\frac{1}{2} * \frac{3}{2}\)
When n = 3, we get \(\frac{2}{3} * \frac{4}{3}\)
When n = 4, we get \(\frac{3}{4} * \frac{5}{4}\)
So overall product is \(\frac{1}{2} * \frac{3}{2} * \frac{2}{3} * \frac{4}{3} * \frac{3}{4} * \frac{5}{4} *... * \frac{19}{20} * \frac{21}{20}\)
Note that all terms are getting cancelled except the first and the last ones.
So we will be left with \(\frac{1}{2} * \frac{21}{20} = \frac{21}{40}\)
Answer (C)
General Discussion
22 Nov 2023, 15:54
Kudos
Bookmarks
I solved this using the formula for : (1-1/2^2)(1-1/3^2)((1-1/4^2)...(1-1/n^2) = n+1/2n. However, I am struggling to understand the logic and how to solve this w/o the need to memorise this formula. I appreciate an explanation.
