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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.­­
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@GMATNinjaGiven: For all positive integers \(r\), the sequence \(S_r\) is given by the formula

\(S_r = \frac{{1}}{{r+1}}-\frac{1}{r}\)­
Asked: What is the sum of the first 20 terms of the sequence?
The first 20 terms of the sequence = \(S_1 + S_2 + ..... + S_{20} = (\frac{1}{2} - \frac{1}{1}) + (\frac{1}{3} - \frac{1}{2}) + .. . + (\frac{1}{21} - \frac{1}{20}) = \frac{1}{21} - \frac{1}{1} = \frac{(1 - 21)}{21} = -\frac{20}{21}\)

​​​​​​​IMO B
­
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