Sequence and Series – Telescoping Sum – JEE Main 28 Jan 2025 Shift 2

Solution:

Given $4a_k = k^2 + 5k + 6 = (k+2)(k+3)$.

So, $\frac{1}{a_k} = \frac{4}{(k+2)(k+3)}$.

We can write using partial fractions:
$$\frac{4}{(k+2)(k+3)} = 4 \left( \frac{1}{k+2} – \frac{1}{k+3} \right)$$

Now, find the sum $S_n$:
$$S_n = \sum_{k=1}^{n} \frac{1}{a_k} = 4 \sum_{k=1}^{n} \left( \frac{1}{k+2} – \frac{1}{k+3} \right)$$

Expand the series:
$$S_n = 4 \left[ \left(\frac{1}{3} – \frac{1}{4}\right) + \left(\frac{1}{4} – \frac{1}{5}\right) + \dots + \left(\frac{1}{n+2} – \frac{1}{n+3}\right) \right]$$

This is a telescoping sum. All terms cancel except the first and the last:
$$S_n = 4 \left( \frac{1}{3} – \frac{1}{n+3} \right) = 4 \left( \frac{n+3-3}{3(n+3)} \right) = \frac{4n}{3(n+3)}$$

We need to find $507 S_{2025}$.
Substitute $n = 2025$:
$$S_{2025} = \frac{4(2025)}{3(2025+3)} = \frac{4 \times 2025}{3 \times 2028}$$

$$507 S_{2025} = \frac{507 \times 4 \times 2025}{3 \times 2028} = \frac{2028 \times 2025}{3 \times 2028} = \frac{2025}{3}$$

$$= 675$$

Ans. (3)