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A series is defined, for positive integer n, by an=n+1n+3−nn+2 What

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A series is defined, for positive integer n, by an=n+1n+3−nn+2 What  [#permalink]

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New post 23 Oct 2018, 22:03
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A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

67% (02:41) correct 33% (03:41) wrong based on 56 sessions

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Re: A series is defined, for positive integer n, by an=n+1n+3−nn+2 What  [#permalink]

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New post 23 Oct 2018, 23:13
\(a_n\) = \(\frac{n+1}{n+3}\) - \(\frac{n}{n+2}\)

\(a_1\) = \(\frac{2}{4}\) - \(\frac{1}{3}\)

\(a_2\) = \(\frac{3}{5}\) - \(\frac{2}{4}\)

\(a_3\) = \(\frac{4}{6}\) - \(\frac{3}{5}\)

\(a_4\) = \(\frac{5}{7}\) - \(\frac{4}{6}\)
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\(a_{58}\) = \(\frac{59}{61}\) - \(\frac{58}{60}\)

\(a_{59}\) = \(\frac{60}{62}\) - \(\frac{59}{61}\)

\(a_{60}\) = \(\frac{61}{63}\) - \(\frac{60}{62}\)

All the terms get canceled out and only \(\frac{61}{63}\) - \(\frac{1}{3}\) remains

\(\frac{61}{63}\) - \(\frac{1}{3}\) = \(\frac{40}{63}\)

OPTION : D
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Re: A series is defined, for positive integer n, by an=n+1n+3−nn+2 What  [#permalink]

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New post 24 Oct 2018, 03:49

Solution


Given:
    • In a series, \(a_n = \frac{(n + 1)}{(n + 3)} – \frac{n}{(n + 2)}\)

To find:
    • The sum of the first 60 terms of this series

Approach and Working:
    • \(a_n = \frac{(n + 1)}{(n + 3)} – \frac{n}{(n + 2)} = \frac{[(n + 1)(n + 2) – n(n + 3)]}{(n + 3)(n + 2)} = \frac{2}{(n + 3)(n + 2)} = 2[\frac{1}{(n + 2)} – \frac{1}{(n + 3)}]\)
    • Thus,
      o \(a_1 = 2(\frac{1}{3} – \frac{1}{4})\)
      o \(a_2 = 2(\frac{1}{4} – \frac{1}{5})\)
      o \(a_3 = 2(\frac{1}{5} – \frac{1}{6})\)
      o .
      o .
      o .
      o \(a_{59} = 2(\frac{1}{61} – \frac{1}{62})\)
      o \(a_{60} = 2(\frac{1}{62} – \frac{1}{63})\)
    • Therefore, \(S = 2(\frac{1}{3} – \frac{1}{4} + \frac{1}{4} – \frac{1}{5} + \frac{1}{5} – \frac{1}{6} + …... + \frac{1}{61} – \frac{1}{62} + \frac{1}{62} – \frac{1}{63}) = 2(\frac{1}{3} – \frac{1}{63}) = 2 * \frac{20}{63} = \frac{40}{63}\)

Hence, the correct answer is Option D

Answer: D

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Re: A series is defined, for positive integer n, by an=n+1n+3−nn+2 What   [#permalink] 24 Oct 2018, 03:49
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