Question

If \( t_n = \dfrac{1}{n(n+2)} \), \( n \in \mathbb{N} \), then which one of the following is true?
Assertion (A): \[ t_1 + t_2 + \cdots + t_{2003} = \dfrac{2003}{3005} \]
Reason (R): \[ t_n = \dfrac{1}{n(n+2)} = \dfrac{1}{2} \left( \dfrac{1}{n} - \dfrac{1}{n+2} \right) \]

• (A) and (R) are true and (R) is a correct explanation of (A)
• (A) and (R) are true, but (R) is not the correct explanation of (A)
• (A) is true, (R) is false
• (A) is false, (R) is true

Hint:

When dealing with rational expressions in series, try to convert the terms using partial fractions. Many such series become telescoping and allow you to cancel intermediate terms, simplifying the summation significantly.

Answer 1

The Correct Option is D

We are given: \[ t_n = \dfrac{1}{n(n+2)} \] Using partial fractions: \[ t_n = \dfrac{1}{n(n+2)} = \dfrac{1}{2} \left( \dfrac{1}{n} - \dfrac{1}{n+2} \right) \] Now, summing from \( n = 1 \) to \( 2003 \): \[ \sum_{n=1}^{2003} t_n = \sum_{n=1}^{2003} \dfrac{1}{2} \left( \dfrac{1}{n} - \dfrac{1}{n+2} \right) \] Factor out \( \dfrac{1}{2} \): \[ = \dfrac{1}{2} \sum_{n=1}^{2003} \left( \dfrac{1}{n} - \dfrac{1}{n+2} \right) \] This is a telescoping series. Write out a few terms to see the cancellation: \[ = \dfrac{1}{2} \left( \left( \dfrac{1}{1} - \dfrac{1}{3} \right) + \left( \dfrac{1}{2} - \dfrac{1}{4} \right) + \left( \dfrac{1}{3} - \dfrac{1}{5} \right) + \cdots + \left( \dfrac{1}{2003} - \dfrac{1}{2005} \right) \right) \] On simplifying, most terms cancel. The remaining terms are: \[ = \dfrac{1}{2} \left( \dfrac{1}{1} + \dfrac{1}{2} - \dfrac{1}{2004} - \dfrac{1}{2005} \right) \] \[ = \dfrac{1}{2} \left( \dfrac{3}{2} - \dfrac{1}{2004} - \dfrac{1}{2005} \right) \] Clearly, this is not equal to \( \dfrac{2003}{3005} \). Therefore, Assertion (A) is false. But Reason (R) is a correct identity. So Reason is true.