Question

Consider a sequence where the nth term, \( t_n = \frac{n}{n+2} \), \( n = 1, 2, \dots \). The value of \( t_3 \times t_4 \times t_5 \times \dots \times t_{53} \) equals:

• A. 2/495
• B. 2/477
• C. 12/55
• D. 1/1485
• E. 1/2970

Hint:

In a telescoping product \( \frac{n}{n+k} \), the number of terms remaining at the start and end is equal to the difference \( k \). Here \( k=2 \), so two terms remain on top and two on the bottom.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This is a "telescoping product" problem. In such sequences, most of the terms in the numerator and denominator cancel each other out, leaving only a few terms at the beginning and the end.
Step 2: Key Formula or Approach:
Write out the first few and last few terms of the product to identify the cancellation pattern:
\[ P = \prod_{n=3}^{53} \frac{n}{n+2} \]
Step 3: Detailed Explanation:
Let's write the expansion: \[ P = \left( \frac{3}{5} \right) \times \left( \frac{4}{6} \right) \times \left( \frac{5}{7} \right) \times \left( \frac{6}{8} \right) \times \dots \times \left( \frac{51}{53} \right) \times \left( \frac{52}{54} \right) \times \left( \frac{53}{55} \right) \] Observation of the pattern:
The denominator of the first term (5) cancels with the numerator of the third term (5).
The denominator of the second term (6) cancels with the numerator of the fourth term (6).
This continues until all numerators from 5 to 53 are cancelled by previous denominators.
We are left with the first two numerators (3 and 4) and the last two denominators (54 and 55).
Calculation: \[ P = \frac{3 \times 4}{54 \times 55} \] \[ P = \frac{12}{2970} \] Divide both by 6: \[ P = \frac{2}{495} \]
Step 4: Final Answer:
The product equals 2/495.