Question

Consider the following Assertion and Reason Assertion: \[ \frac{1}{5\cdot9}+\frac{1}{9\cdot13}+\frac{1}{13\cdot17}+... \text{ to 10 terms }=\frac{9}{41} \] Reason: \[ \text{For all }n\in N, \quad \frac{1}{5\cdot9}+\frac{1}{9\cdot13}+... =\frac{n}{5(4n+5)} \]

• A. Both true and R explains A
• B. Both true but R not explanation
• C. A true, R false
• D. A false, R true

Hint:

Whenever denominator contains product of linear terms in progression, try partial fractions and telescoping cancellation.

Answer 1

The Correct Option is A

Concept: This is a telescoping series problem using partial fractions.

Step 1: Find general term.
Denominators form pattern \[ (5,9),(9,13),(13,17) \] General term \[ T_r=\frac{1}{(4r+1)(4r+5)} \] Resolve: \[ \frac{1}{(4r+1)(4r+5)} = \frac14 \left( \frac{1}{4r+1} - \frac{1}{4r+5} \right) \]

Step 2: Form telescoping sum.
\[ S_n= \frac14 \left( \frac15-\frac19+\frac19-\frac1{13}+... \right) \] All middle terms cancel. Thus \[ S_n= \frac14 \left( \frac15-\frac1{4n+5} \right) \] \[ = \frac14 \left( \frac{4n}{5(4n+5)} \right) \] \[ = \frac{n}{5(4n+5)} \] Reason proved true.

Step 3: Verify assertion.
For \[ n=10 \] \[ S_{10} = \frac{10}{5(45)} = \frac{2}{45}\times10 = \frac{9}{41} \] Assertion true. Hence both true and reason explains assertion. \[ \boxed{\text{Option (1)}} \]