Sequences & Series
Sequence and Series
star_batch_jee_advanced_2025
Grade 11
Question:
The sum $\sum_{k=1}^{n} \frac{k^2 - \frac{1}{2}}{k^4 + \frac{1}{4}}$ is equal to:
\frac{2n^2 - 2n + 1}{2n^2 + 2n + 1}
\frac{2n^2 - n}{2n^2 + 2n + 1}
\frac{n^2}{2n^2 + 2n + 1}
\frac{2n^2}{2n^2 + 2n + 1}
Step-by-Step Solution
Key Concept: Partial fraction decomposition reveals a telescoping series where consecutive terms cancel.
The sum $\sum_{k=1}^{n} \frac{k^2 - \frac{1}{2}}{(k^2 - k + 1)(k^2 + k + 1)}$ is rewritten using partial fractions as $\sum_{k=1}^{n} \left(\frac{k - \frac{1}{2}}{k^2 - k + \frac{1}{2}} - \frac{k + \frac{1}{2}}{k^2 + k + \frac{1}{2}}\right)$. This telescopes to $1 - \frac{n + \frac{1}{2}}{n^2 + n + 1} = 1 - \frac{2n+1}{2n^2 + 2n + 1}$.
Correct Answer: 4