Mauryashristi61
What is the approach to this one ?
For all positive integers \(r\), the sequence \(S_r\) is given by the formula\(S_r = \frac{{1}}{{r+1}}-\frac{1}{r}\)
What is the sum of the first 20 terms of the sequence?(A) \(\frac{-22}{21}\)
(B) \(\frac{-20}{21}\)
(C) \(\frac{1}{21}\)
(D) \(\frac{20}{21}\)
(E) \(\frac{22}{21}\)
You can check GMATNinja's explaination in the video above. However, here it is again.
Since \(S_r = \frac{{1}}{{r+1}}-\frac{1}{r}\), then:
\(s_1 = \frac{1}{1+1} - \frac{1}{1} = \frac{1}{2} - 1\)
\(s_2 = \frac{1}{2+1} - \frac{1}{2} = \frac{1}{3} - \frac{1}{2}\)
\(s_3 = \frac{1}{3+1} - \frac{1}{3} = \frac{1}{4} - \frac{1}{3}\)
...
\(s_{19} = \frac{1}{19+1} - \frac{1}{19} = \frac{1}{20} - \frac{1}{19}\)
\(s_{20} = \frac{1}{20+1} - \frac{1}{20} = \frac{1}{21} - \frac{1}{20}\)
Therefore, the sum of the first 20 terms of the sequence is:
\(( \frac{1}{2} - 1) + ( \frac{1}{3} - \frac{1}{2}) + (\frac{1}{4} - \frac{1}{3}) + ... + (\frac{1}{20} - \frac{1}{19}) + (\frac{1}{21} - \frac{1}{20})= \)
\(= - 1 + \frac{1}{21} = \)
\(=- \frac{20}{21} \)
Answer: B.