Sequences & Series
Harmonic-Type Series Difference — Finding α
nta_pyq_2024_apr
Grade None
Question:
If $\left(\dfrac{1}{\alpha+1}+\dfrac{1}{\alpha+2}+\cdots+\dfrac{1}{\alpha+1012}\right)-\left(\dfrac{1}{2\cdot1}+\dfrac{1}{4\cdot3}+\dfrac{1}{6\cdot5}+\cdots+\dfrac{1}{2024\cdot2023}\right)=\dfrac{1}{2024}$, then $\alpha$ is equal to
Step-by-Step Solution
Key Concept: Second sum: $\sum_{k=1}^{1012}\frac{1}{2k(2k-1)}=\sum_{k=1}^{1012}\left(\frac{1}{2k-1}-\frac{1}{2k}\right)$. First sum: $\sum_{k=1}^{1012}\frac{1}{\alpha+k}$. Set the difference equal to $1/2024$.
$\alpha=1011$.
Correct Answer: 1011