Sequences & Series
Telescoping sum — reduced fraction components
MJAT_TS4_P2
Grade 12
Question:
Let $x=\displaystyle\sum_{n=1}^{2025}\frac{2n^2-1}{n^2(n+1)^2}$. The value of $x$ is $\dfrac{\alpha}{\beta}$ (in reduced form). Then $\alpha+\beta=$
Step-by-Step Solution
Key Concept: Decompose: $\frac{2n^2-1}{n^2(n+1)^2}=\frac{2n-1}{n^2}-\frac{2n+1}{(n+1)^2}$. This telescopes: $\sum_{n=1}^N\left[\frac{2n-1}{n^2}-\frac{2n+1}{(n+1)^2}\right]=\frac{1}{1}-\frac{2N+1}{(N+1)^2}$.
$x=(2025/2026)^2$. With $\alpha=2025$, $\beta=2026$ (bases of the squared fraction): $\alpha+\beta=\mathbf{4051}$.
Correct Answer: 4051