Question

\(\frac{1}{3.6} + \frac{1}{6.9} + \frac{1}{9.12} + \dots\) up to 9 terms \(=\)

• A. \(\tfrac{10}{99}\)
• B. \(\tfrac{11}{108}\)
• C. \(\tfrac{1}{10}\)
• D. \(\tfrac{1}{90}\)

Hint:

Telescoping series often simplify dramatically by partial fraction decomposition.

Answer 1

The Correct Option is C

Step 1: Observe each term.
Each term has the form \(\frac{1}{(3k)\,(3k+3)}\) where \(k = 1,2,3,\dots\). For instance: \[ \frac{1}{3 \cdot 6},\quad \frac{1}{6 \cdot 9},\quad \frac{1}{9 \cdot 12},\dots \] Generally, \[ \frac{1}{(3k)\,[3(k+1)]} = \frac{1}{3\cdot 3} \cdot \frac{1}{k(k+1)} = \frac{1}{9}\Bigl[\frac{1}{k} - \frac{1}{k+1}\Bigr]. \] Step 2: Summation of 9 terms.
When we expand: \[ \sum_{k=1}^{9} \frac{1}{(3k)(3k+3)} = \frac{1}{9} \sum_{k=1}^{9} \Bigl(\frac{1}{k} - \frac{1}{k+1}\Bigr) = \frac{1}{9} \Bigl(1 - \frac{1}{10}\Bigr) = \frac{1}{9} \cdot \frac{9}{10} = \frac{1}{10}. \] Hence, the sum up to 9 terms is \(\boxed{\frac{1}{10}}\).