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23 Nov 2023, 02:30
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B
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Difficulty:
95%
(hard)
Question Stats:
37% (03:01) correct
63%
(02:17)
wrong
based on 62
sessions
History
Date
Time
Result
Not Attempted Yet
The \(n_{th}\) term of a sequence \(a_1, \ a_2, \ a_3 \ … \ a_n\) is given by \(a_n = \frac{1}{n(n+2)}\). What is the sum of the first 20 terms of the sequence?
(A) 300/440
(B) 325/462
(C) 303/462
(D) 375/450
(E) 650/462
(A) 300/440
(B) 325/462
(C) 303/462
(D) 375/450
(E) 650/462
24 Nov 2023, 10:34
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Tn = \(\frac{1}{n(n+2)}\) = \(\frac{[m][fraction]1}{n} \)- \(\frac{1}{(n+2)}\)/2[/fraction][/m]
2Sn = \(\frac{1}{1.3}\) + \(\frac{1}{2.4}\)+ \(\frac{1}{3.5}\)...........................\(\frac{1}{18.20}\) + \(\frac{1}{19.21}\) + \(\frac{1}{20.22}\)
2Sn = \(\frac{1}{1}\) - \(\frac{1}{3}\) + \(\frac{1}{2}\) - \(\frac{1}{4}\) + \(\frac{1}{3}\) - \(\frac{1}{5}\) ........................\(\frac{1}{18}\) - \(\frac{1}{20}\) + \(\frac{1}{19}\)- \(\frac{1}{21}\) + \(\frac{1}{20}\) - \(\frac{1}{22}\)
2Sn = 1 + \(\frac{1}{2}\) - \(\frac{1}{21}\) - \(\frac{1}{22}\) = \(\frac{650}{462}\)
Sn = \(\frac{325}{462}\)
O\(\)ption B
2Sn = \(\frac{1}{1.3}\) + \(\frac{1}{2.4}\)+ \(\frac{1}{3.5}\)...........................\(\frac{1}{18.20}\) + \(\frac{1}{19.21}\) + \(\frac{1}{20.22}\)
2Sn = \(\frac{1}{1}\) - \(\frac{1}{3}\) + \(\frac{1}{2}\) - \(\frac{1}{4}\) + \(\frac{1}{3}\) - \(\frac{1}{5}\) ........................\(\frac{1}{18}\) - \(\frac{1}{20}\) + \(\frac{1}{19}\)- \(\frac{1}{21}\) + \(\frac{1}{20}\) - \(\frac{1}{22}\)
2Sn = 1 + \(\frac{1}{2}\) - \(\frac{1}{21}\) - \(\frac{1}{22}\) = \(\frac{650}{462}\)
Sn = \(\frac{325}{462}\)
O\(\)ption B
28 Nov 2023, 07:51
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sj296
Hey, can you help me understand how you moved to 2Sn and also the first line is a bit confusing. Would really appreciate your help on this. TIA
