Question

The sum \[ \sum_{n=4}^{\infty} \frac{6}{n^2 - 4n + 3} \quad \text{equals} \]

• A. \( \frac{5}{2} \)
• B. 3
• C. \( \frac{7}{2} \)
• D. \( \frac{9}{2} \)

Hint:

In a telescoping series, most terms cancel out, leaving only a few terms that do not have a counterpart.

Answer 1

The Correct Option is D

Step 1: Simplify the series expression.
We start by factoring the denominator in the given sum: \[ n^2 - 4n + 3 = (n-1)(n-3) \] Thus, the general term of the sum becomes: \[ \frac{6}{(n-1)(n-3)} \] We can now apply partial fraction decomposition: \[ \frac{6}{(n-1)(n-3)} = \frac{A}{n-1} + \frac{B}{n-3} \] Solving for \(A\) and \(B\), we find: \[ A = 3, \quad B = -3 \] Hence, the sum becomes: \[ \sum_{n=4}^{\infty} \left( \frac{3}{n-1} - \frac{3}{n-3} \right) \] This is a telescoping series, and most terms cancel out, leaving: \[ \text{Final result: } 3 \]