<p>315. The sum of infinite series \(\frac{1}{1 \cdot 4} + \frac{1}{4 \cdot 7} + \frac{1}{7 \cdot 10} + \ldots\) to \(\infty\) is</p>
Step-by-Step Solution
Key Concept: Recognize that consecutive denominators form an arithmetic sequence with common difference 3, then use partial fractions decomposition: 1/[n(n+3)] = (1/3)[1/n - 1/(n+3)] to create a telescoping series.
<p><strong>Step 1:</strong> Identify the pattern. The denominators are 1·4, 4·7, 7·10, ... which can be written as n(n+3) where n = 1, 4, 7, ... (i.e., n = 3k-2 for k = 1, 2, 3, ...)</p><p><strong>Step 2:</strong> Apply partial fraction decomposition:</p><p>$$\frac{1}{n(n+3)} = \frac{1}{3}\left(\frac{1}{n} - \frac{1}{n+3}\right)$$</p><p><strong>Step 3:</strong> Write out the series with the first few terms:</p><p>$$S = \frac{1}{3}\left[\left(\frac{1}{1} - \frac{1}{4}\right) + \left(\frac{1}{4} - \frac{1}{7}\right) + \left(\frac{1}{7} - \frac{1}{10}\right) + \ldots\right]$$</p><p><strong>Step 4:</strong> Observe the telescoping pattern. The partial sums telescope:</p><p>$$S_n = \frac{1}{3}\left(\frac{1}{1} - \frac{1}{n+3}\right)$$</p><p><strong>Step 5:</strong> As n → ∞, we have 1/(n+3) → 0, so:</p><p>$$S = \frac{1}{3}\left(1 - 0\right) = \frac{1}{3}$$</p><p>∴ Answer: <strong>1/3</strong></p>
Correct Answer: A