Question

The sum of the infinite series \(1 + \frac{5}{6} + \frac{12}{6^2} + \frac{22}{6^3} + \frac{35}{6^4} + \dots\) is equal to:

• A. \( \frac{425}{216} \)
• B. \( \frac{429}{216} \)
• C. \( \frac{288}{125} \)
• D. \( \frac{280}{125} \)

Hint:

Use series transformation and subtraction techniques to find closed forms for complex series.

Answer 1

The Correct Option is C

Step 1: Let the series be defined as: \[ S = 1 + \frac{5}{6} + \frac{12}{6^2} + \frac{22}{6^3} + \frac{35}{6^4} + \cdots \quad {(i)} \] Now, define another series: \[ \frac{S}{6} = 1 + \frac{5}{6^2} + \frac{12}{6^3} + \frac{22}{6^4} + \cdots \quad {(ii)} \] Step 2: On subtracting equation (ii) from equation (i): \[ S - \frac{S}{6} = \left(1 + \frac{5}{6} + \frac{12}{6^2} + \cdots \right) - \left(1 + \frac{5}{6^2} + \frac{12}{6^3} + \cdots \right) \] Step 3: This simplifies to: \[ \frac{5S}{6} = 1 + \frac{4}{6} + \frac{7}{6^2} + \frac{10}{6^3} + \cdots \] Now, multiply both sides by 36: \[ \frac{25S}{36} = 1 + \frac{3}{6} + \frac{3}{6^2} + \frac{3}{6^3} + \cdots \] Step 4: The right-hand side is a geometric series with the first term 1 and the common ratio \( \frac{1}{6} \). The sum of this geometric series is: \[ S = \frac{8}{5} \times \frac{36}{25} = \frac{288}{125} \] Thus, the sum of the series is: \[ S = \frac{288}{125} \] Thus, the value of \( S \) matches option (C).