Let the greatest common divisor of m and n be | vedclass

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(C) The given series is $\sum_{k=0}^{19} \frac{1}{(6k+1)(6k+7)}$.This can be written as $\frac{1}{6} \sum_{k=0}^{19} \left( \frac{1}{6k+1} - \frac{1}{

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$342$

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(C) The given series is $\sum_{k=0}^{19} \frac{1}{(6k+1)(6k+7)}$.This can be written as $\frac{1}{6} \sum_{k=0}^{19} \left( \frac{1}{6k+1} - \frac{1}{6k+7} ight)$.This is a telescoping series:$\frac{1}{6} \left[ (\frac{1}{1} - \frac{1}{7}) + (\frac{1}{7} - \frac{1}{13}) + \dots + (\frac{1}{115} - \frac{1}{121}) ight]$.$= \frac{1}{6} \left( 1 - \frac{1}{121} ight) = \frac{1}{6} 	imes \frac{120}{121} = \frac{20}{121}$.Given $\frac{m}{n} = \frac{20}{121}$,where $\gcd(m, n) = 1$,we have $m = 20$ and $n = 121$.Therefore,$5m + 2n = 5(20) + 2(121) = 100 + 242 = 342$.

Let the greatest common divisor of $m$ and $n$ be $1$. If $\frac{1}{1 \cdot 7} + \frac{1}{7 \cdot 13} + \frac{1}{13 \cdot 19} + \dots$ up to $20$ terms $= \frac{m}{n}$,then $5m + 2n = $

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