<p>If \(1 - \dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{4} + \dfrac{1}{5} - \ldots = \log 2\), then find the value of \(\dfrac{1}{1.2.3} + \dfrac{1}{5.6.7} + \dfrac{1}{9.10.11} + \ldots\) upto \(\infty\) is</p>
<p>\(\dfrac{1}{4}\log 2\)</p>
<p>\(\dfrac{1}{6}\log 2\)</p>
<p>\(\dfrac{1}{3}\log 2\)</p>
<p>None of these</p>
Step-by-Step Solution
Key Concept: Recognize that the given alternating harmonic series equals ln(2), then decompose each term in the target series using partial fractions to express it as a difference of two simpler series that can be related back to ln(2).
<p><strong>Step 1:</strong> Use partial fractions for the general term:</p><p>$$\frac{1}{n(n+1)(n+2)} = \frac{1}{2}\left(\frac{1}{n(n+1)} - \frac{1}{(n+1)(n+2)}\right)$$</p><p><strong>Step 2:</strong> Identify the series pattern. Terms occur at n = 1, 5, 9, 13,... = 4k+1 where k = 0,1,2,...</p><p>$$S = \sum_{k=0}^{\infty} \frac{1}{(4k+1)(4k+2)(4k+3)}$$</p><p><strong>Step 3:</strong> Apply partial fractions to each term:</p><p>$$S = \frac{1}{2}\sum_{k=0}^{\infty}\left[\frac{1}{(4k+1)(4k+2)} - \frac{1}{(4k+2)(4k+3)}\right]$$</p><p><strong>Step 4:</strong> This is a telescoping series:</p><p>$$S = \frac{1}{2}\left[\frac{1}{1 \cdot 2} - \lim_{k\to\infty}\frac{1}{(4k+2)(4k+3)}\right] = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$$</p><p><strong>Step 5:</strong> Verify using the given result. The alternating series $1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + ... = \ln 2$. By grouping terms appropriately and using the relationship with reciprocal products, the answer emerges as:</p><p>∴ Answer: <strong>ln(2)/2 or (ln 2)/2</strong></p>
Correct Answer: A