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Sequence Product Telescoping Sum

XAT 2023 · Quantitative Ability & Data Interpretation · Medium · Sequences and Series

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Consider an+1=11+1ana_{n+1} = \frac{1}{1 + \frac{1}{a_n}} for n=1,2,…,2008,2009n = 1, 2, \dots, 2008, 2009 where a1=1a_1 = 1. Find the value of a1a2+a2a3+a3a4+⋯+a2008a2009a_1 a_2 + a_2 a_3 + a_3 a_4 + \dots + a_{2008} a_{2009}.

  1. A.

    20091000\frac{2009}{1000}

  2. B.

    20092008\frac{2009}{2008}

  3. C.

    20082009\frac{2008}{2009}

  4. D.

    60002009\frac{6000}{2009}

  5. E.

    20086000\frac{2008}{6000}

Answer

C

Explanation

a1=1a_1 = 1, a2=1/2a_2 = 1/2, a3=1/3,…,an=1/na_3 = 1/3, \dots, a_n = 1/n. anan+1=1n(n+1)=1n−1n+1a_n a_{n+1} = \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}. Sum = (1−1/2)+(1/2−1/3)+⋯+(1/2008−1/2009)=1−12009=20082009(1 - 1/2) + (1/2 - 1/3) + \dots + (1/2008 - 1/2009) = 1 - \frac{1}{2009} = \frac{2008}{2009}.

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