<p>Sum to infinite terms of the series \(\dfrac{1}{1.3} + \dfrac{2}{1.3.5} + \dfrac{3}{1.3.5.7} + \dfrac{4}{1.3.5.7.9} + \ldots\) is:</p>
Step-by-Step Solution
Key Concept: Recognize that the denominator of the nth term is the product of the first n odd numbers (2n-1)!!/1, allowing us to express each term using double factorial notation and then telescope the series using partial fractions or a telescoping technique.
<p><strong>Step 1:</strong> Identify the pattern. The nth term has numerator n and denominator is the product of first n odd numbers: $1 \cdot 3 \cdot 5 \cdot 7 \cdots (2n-1) = (2n-1)!!$</p><p><strong>Step 2:</strong> Express the general term: $a_n = \frac{n}{1 \cdot 3 \cdot 5 \cdots (2n-1)}$</p><p><strong>Step 3:</strong> Use the key insight that $\frac{n}{1 \cdot 3 \cdot 5 \cdots (2n-1)} = \frac{1}{1 \cdot 3 \cdot 5 \cdots (2n-3)} - \frac{1}{1 \cdot 3 \cdot 5 \cdots (2n-1)}$ (this can be verified by taking difference of consecutive terms)</p><p><strong>Step 4:</strong> Write out the series: $S = \left(\frac{1}{1} - \frac{1}{1 \cdot 3}\right) + \left(\frac{1}{1 \cdot 3} - \frac{1}{1 \cdot 3 \cdot 5}\right) + \left(\frac{1}{1 \cdot 3 \cdot 5} - \frac{1}{1 \cdot 3 \cdot 5 \cdot 7}\right) + \cdots$</p><p><strong>Step 5:</strong> This is a telescoping series where all intermediate terms cancel: $S = 1 - \lim_{n \to \infty} \frac{1}{1 \cdot 3 \cdot 5 \cdots (2n-1)} = 1 - 0 = 1$</p><p>∴ Answer: B</p>
Correct Answer: B