<s>[INST] Find the sum of $\frac {3}{4} + \frac {3}{28} + \frac {3}{70} + \frac {3}{130} .... + \frac {3}{9700}$ What will be the sum of :
$\frac {3}{4} + \frac {3}{28} + \frac {3}{70} + \frac {3}{130} .... + \frac {3}{9700}$ ?
 [/INST] $\frac {3}{4} + \frac {3}{28} + \frac {3}{70} + \frac {3}{130} .... + \frac {3}{9700}=\frac {4-1}{4} + \frac {7-4}{28} + \frac {10-7}{70} + \frac {13-10}{130} .... + \frac {100-97}{9700}=\left(\frac {4}{4} + \frac {7}{28} + \frac {10}{70} + \frac {13}{130} .... + \frac {100}{9700}\right)-\left( \frac {1}{4} + \frac {4}{28} + \frac {7}{70} + \frac {10}{130} .... + \frac {97}{9700} \right)=1-\frac{1}{100}=\frac{99}{100}$
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