Sequences & Series
Partial Fraction Telescoping for Reciprocal Sum
nta_pyq_2025_apr
Grade 11
Question:
Let $4a_n = n^2+5n+6$ and $S_n = \displaystyle\sum_{k=1}^n\dfrac{1}{a_k}$. Then $507\,S_{2025}$ equals
Step-by-Step Solution
Key Concept: Factor $n^2+5n+6=(n+2)(n+3)$ so that $1/a_k=4/[(k+2)(k+3)]$ telescopes via partial fractions.
$a_k=\dfrac{(k+2)(k+3)}{4}$, so $\dfrac{1}{a_k}=\dfrac{4}{(k+2)(k+3)}=4\left(\dfrac{1}{k+2}-\dfrac{1}{k+3}\right)$.
$$S_n=4\left(\frac{1}{3}-\frac{1}{n+3}\right)=\frac{4n}{3(n+3)}.$$
$$S_{2025}=\frac{4\times2025}{3\times2028}=\frac{8100}{6084}=\frac{675}{507}.$$
$$507\,S_{2025}=675.$$
Correct Answer: 2