Sequences & Series
Sequence and Series
star_batch_jee_advanced_2025
Grade 11
Question:
$\frac{n}{1 \cdot 2 \cdot 3} + \frac{n-1}{2 \cdot 3 \cdot 4} + \frac{n-2}{3 \cdot 4 \cdot 5} + \ldots$ upto $n$ terms is equal to:
\frac{1}{2(n+2)} - \frac{n+1}{4} - \frac{1}{2}
\frac{1}{2(n+2)} + \frac{n+1}{4} - \frac{1}{2}
\frac{1}{n+2} + \frac{n+1}{4} - \frac{1}{2}
\frac{1}{2(n+2)} + \frac{1}{2} + \frac{1}{2}
Step-by-Step Solution
Key Concept: Use partial fractions and telescoping sums to simplify products of consecutive integers in denominators.
The sum $\sum_{r=1}^{n} rac{n-(r-1)}{r(r+1)(r+2)} = rac{n+1}{2} \sum_{r=1}^{n} \left( rac{1}{r(r+1)} - rac{1}{(r+1)(r+2)}
ight)$. This telescopes to give $ rac{n+1}{2}\left( rac{1}{2} - rac{1}{n+2}
ight) = rac{n+1}{4} - rac{1}{2(n+2)} = rac{n(n+1)}{4(n+2)}$.
Correct Answer: 1