Sequences & Series
Telescoping series with partial fractions
nta_pyq_2023_jan
Grade 11
Question:
If $a_n = \dfrac{-2}{4n^2 - 16n + 15}$, then $a_1 + a_2 + \ldots + a_{25}$ is equal to:
\dfrac{51}{144}
\dfrac{49}{138}
\dfrac{50}{141}
\dfrac{52}{147}
Step-by-Step Solution
Key Concept: Factor the denominator: $4n^2 - 16n + 15 = (2n-3)(2n-5)$. Use partial fractions: $\frac{-2}{(2n-3)(2n-5)} = \frac{1}{2n-3} - \frac{1}{2n-5}$ (telescoping).
$\sum_{n=1}^{25} a_n = \sum \left(\frac{1}{2n-3} - \frac{1}{2n-5}\right) = \frac{1}{47} - \frac{1}{-3} = \frac{1}{47} + \frac{1}{3} = \frac{3+47}{141} = \frac{50}{141}$.
Correct Answer: 3