Question

The sum of the series \( \frac{1}{1.5} + \frac{1}{5.9} + \frac{1}{9.13} + \cdots \) up to \( n \) terms is:

• A. \( \frac{1}{4n + 1} \)
• B. \( \frac{4}{4n + 1} \)
• C. \( \frac{n}{4n + 1} \)
• D. \( \frac{4n + 1}{5(4n + 1)} \)

Hint:

For series where the terms follow a specific pattern, identify the general form of the denominator and simplify the sum accordingly.

Answer 1

The Correct Option is C

We are given the series: \[ S_n = \frac{1}{1.5} + \frac{1}{5.9} + \frac{1}{9.13} + \cdots \quad \text{up to \( n \) terms}. \] First, observe the pattern in the denominators. The terms in the denominators follow a sequence of the form: \[ 1.5, 5.9, 9.13, \cdots \] This suggests that the general form of the denominator for the \( k \)-th term is: \[ (4k - 2) + 0.5 \] Thus, the \( k \)-th term of the series can be written as: \[ T_k = \frac{1}{(4k - 2) + 0.5} = \frac{1}{4k + 1}. \] Step 1: The sum of the series up to \( n \) terms is: \[ S_n = \sum_{k=1}^{n} \frac{1}{4k + 1}. \] This simplifies to: \[ S_n = \frac{1}{4(1) + 1} + \frac{1}{4(2) + 1} + \frac{1}{4(3) + 1} + \cdots + \frac{1}{4n + 1}. \] 
Step 2: The general formula for the sum of this series up to \( n \) terms is: \[ S_n = \frac{n}{4n + 1}. \] Thus, the sum of the series is \( \frac{n}{4n + 1} \).