Question

Consider a sequence \( S \) whose \( n \)th term \( T_n \) is defined as \( T_n = 1 + \frac{3}{n} \), where \( n = 1, 2, \ldots \). Find the product of all the consecutive terms of \( S \) starting from the 4th term to the 60th term.

• A. 1980.55
• B. 1985.55
• C. 1990.55
• D. 1975.55

Hint:

Telescoping products often reduce to factorial expressions — cancel terms before computing.

Answer 1

The Correct Option is A

Given \( T_n = 1 + \frac{3}{n} = \frac{n+3}{n} \)
We are to calculate: \[ \prod_{n=4}^{60} T_n = \prod_{n=4}^{60} \frac{n+3}{n} \] \[ = \frac{7}{4} \cdot \frac{8}{5} \cdot \frac{9}{6} \cdots \frac{63}{60} \] This is a telescoping product: \[ = \frac{7 \cdot 8 \cdot \ldots \cdot 63}{4 \cdot 5 \cdot \ldots \cdot 60} = \frac{63! / 6!}{60! / 3!} = \frac{63! \cdot 3!}{60! \cdot 6!} \] Evaluating: \[ \frac{63 \cdot 62 \cdot 61 \cdot 3!}{6!} = \frac{63 \cdot 62 \cdot 61 \cdot 6}{720} \approx \boxed{1980.55} \]