EgmatQuantExpert wrote:

e-GMAT Question of the Week #37The \(n^{th}\) 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. \(\frac{300}{440}\)

B. \(\frac{325}{462}\)

C. \(\frac{303}{462}\)

D. \(\frac{375}{450}\)

E. \(\frac{650}{462}\)

\(a_n = \frac{1}{n(n+2)}\)

\(a_1 = \frac{1}{1(1+2)} = \frac{1}{1*3} = \frac{1}{2}[(\frac{1}{1}) - (\frac{1}{3})]\)

\(a_2 = \frac{1}{2(2+2)} = \frac{1}{2*4}= \frac{1}{2}[(\frac{1}{2}) - (\frac{1}{4})]\)

\(a_3 = \frac{1}{3(3+2)} = \frac{1}{3*5}= \frac{1}{2}[(\frac{1}{3}) - (\frac{1}{5})]\)

\(a_4 = \frac{1}{4(4+2)} = \frac{1}{4*6}= \frac{1}{2}[(\frac{1}{4}) - (\frac{1}{6})]\)

\(a_5 = \frac{1}{5(5+2)} = \frac{1}{5*7}= \frac{1}{2}[(\frac{1}{5}) - (\frac{1}{7})]\)

and so on....

\(a_{20} = \frac{1}{20(20+2)} = \frac{1}{20*22}= \frac{1}{2}[(\frac{1}{20}) - (\frac{1}{22})]\)

Adding them we can see that all terms will be cancelled out except

Sum of 20 terms \(= (\frac{1}{2})[ (\frac{1}{1}) - \frac{1}{3} + (\frac{1}{2}) - \frac{1}{4} + \frac{1}{3} - \frac{1}{5} + \frac{1}{4} - \frac{1}{6} + \frac{1}{5} - \frac{1}{7} + ------- +\frac{1}{16} - \frac{1}{18} + \frac{1}{17} - \frac{1}{19} + \frac{1}{18} - \frac{1}{20} + \frac{1}{19} - (\frac{1}{21}) + \frac{1}{20} - (\frac{1}{22}) ]\)

Sum of 20 terms \(= (\frac{1}{2})[ (\frac{1}{1}) + (\frac{1}{2}) - (\frac{1}{21}) - (\frac{1}{22}) ]\)

Sum of 20 terms = \(\frac{325}{462}\)

Answer: Option B

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