Sequences & Series
Sequences and Series
nta_pyq_2025_jan
Grade 11
Question:
For positive integers $n$, if $4a_{n}=(n^{2}+5n+6)$ and $S_{n}=\displaystyle\sum_{k=1}^{n}\dfrac{1}{a_{k}}$, then the value of $507\,S_{2025}$ is:
Step-by-Step Solution
Key Concept: $n^{2}+5n+6=(n+2)(n+3)$ factors cleanly, and $\dfrac{1}{(n+2)(n+3)}=\dfrac{1}{n+2}-\dfrac{1}{n+3}$ telescopes.
$4a_{n}=(n+2)(n+3)\Rightarrow \dfrac{1}{a_{n}}=\dfrac{4}{(n+2)(n+3)}=4\!\left[\dfrac{1}{n+2}-\dfrac{1}{n+3}\right].$
$$S_{n}=4\sum_{k=1}^{n}\!\left[\dfrac{1}{k+2}-\dfrac{1}{k+3}\right]=4\!\left[\dfrac{1}{3}-\dfrac{1}{n+3}\right].$$
$$S_{2025}=4\!\left[\dfrac{1}{3}-\dfrac{1}{2028}\right]=4\cdot\dfrac{675}{3\cdot 2028}=\dfrac{2700}{2028}=\dfrac{675}{507}.$$
$$507\,S_{2025}=675.$$
Correct Answer: 2