Question

Evaluate \[ \frac{1}{1\cdot6}+\frac{1}{6\cdot11}+\frac{1}{11\cdot16}+\cdots \] for \(n\) terms.

• A. \(\dfrac{n}{5n+1}\)
• B. \(\dfrac{5}{5n+1}\)
• C. \(\dfrac{5n}{5n+1}\)
• D. \(\dfrac{n}{5(5n+1)}\)

Hint:

Whenever the denominator is of the form \[ (a)(a+k), \] use \[ \boxed{ \frac1{a(a+k)} = \frac1k \left( \frac1a-\frac1{a+k} \right) } \] which usually produces a telescoping series.

Answer 1

The Correct Option is A

Step 1: Write the general term. The denominators are \[ (1,6),\ (6,11),\ (11,16),\ldots \] Hence, the \(r^{\text{th}}\) term is \[ T_r=\frac{1}{(5r-4)(5r+1)}, \qquad r=1,2,\ldots,n. \] Therefore, \[ S_n=\sum_{r=1}^{n}\frac{1}{(5r-4)(5r+1)}. \]

Step 2: Resolve into partial fractions. Let \[ \frac{1}{(5r-4)(5r+1)} = \frac{A}{5r-4} +\frac{B}{5r+1}. \] Then, \[ 1=A(5r+1)+B(5r-4). \] Solving, \[ A=\frac15,\qquad B=-\frac15. \] Hence, \[ \frac{1}{(5r-4)(5r+1)} = \frac15 \left( \frac1{5r-4} - \frac1{5r+1} \right). \]

Step 3: Apply telescoping. Thus, \[ S_n = \frac15 \sum_{r=1}^{n} \left( \frac1{5r-4} - \frac1{5r+1} \right). \] Expanding, \[ S_n = \frac15 \left[ \left(1-\frac16\right) + \left(\frac16-\frac1{11}\right) +\cdots+ \left(\frac1{5n-4}-\frac1{5n+1}\right) \right]. \] All intermediate terms cancel. Hence, \[ S_n = \frac15 \left( 1-\frac1{5n+1} \right). \] \[ = \frac15 \cdot \frac{5n}{5n+1} = \boxed{\frac{n}{5n+1}}. \] Hence, \[ \boxed{(A)} \] is the correct answer.