frac{1}{3 cdot 6} + frac{1}{6 cdot 9} + frac{ | vedclass

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(C) Let $S = \frac{1}{3 \cdot 6} + \frac{1}{6 \cdot 9} + \frac{1}{9 \cdot 12} + \dots$ to $9$ terms. The $n$-th term is $T_n = \frac{1}{(3n)(3n+3)} =

Vedclass · Accepted Answer

$\frac{1}{10}$

Vedclass · Answer

(C) Let $S = \frac{1}{3 \cdot 6} + \frac{1}{6 \cdot 9} + \frac{1}{9 \cdot 12} + \dots$ to $9$ terms. The $n$-th term is $T_n = \frac{1}{(3n)(3n+3)} = \frac{1}{9n(n+1)}$. We can write $T_n = \frac{1}{9} \left( \frac{1}{n} - \frac{1}{n+1} \right)$. Summing for $n=1$ to $9$: $S = \frac{1}{9} \sum_{n=1}^{9} \left( \frac{1}{n} - \frac{1}{n+1} \right)$. This is a telescoping series: $S = \frac{1}{9} \left[ (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + \dots + (\frac{1}{9} - \frac{1}{10}) \right]$. $S = \frac{1}{9} \left( 1 - \frac{1}{10} \right) = \frac{1}{9} \left( \frac{9}{10} \right) = \frac{1}{10}$.

$\frac{1}{3 \cdot 6} + \frac{1}{6 \cdot 9} + \frac{1}{9 \cdot 12} + \dots$ to $9$ terms $=$

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