Question

Consider a sequence where the $n$th 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. $\frac{2}{495}$
• B. $\frac{2}{477}$
• C. $\frac{12}{55}$
• D. $\frac{1}{1485}$
• E. $\frac{1}{2970}$

Hint:

In telescoping products, most terms cancel out — always check for patterns between successive numerators and denominators.

Answer 1

The Correct Option is A

We have: \[ t_n = \frac{n}{n+2} \] The product from $t_3$ to $t_{53}$ is: \[ \prod_{n=3}^{53} \frac{n}{n+2} \] This telescopes as: \[ \frac{3}{5} \times \frac{4}{6} \times \frac{5}{7} \times \dots \times \frac{53}{55} \] Canceling common terms from numerator and denominator, we are left with: \[ \frac{3 \times 4}{54 \times 55} = \frac{12}{2970} = \frac{2}{495} \] \[ \boxed{\frac{2}{495}} \]