If ((1)/(α+1)+(1)/(α+2)+ +(1)/(α+1012))-((1)/(2 · 1)+(1)/(4 · 3)+(1)/(6 · 5)+ . .+(1)/(2024 · 2023))=(1)/(2024) , then α is equal to________

Correct answer: 0 aligned & ((1)/(α+1)+(1)/(α+2)+ +(1)/(α+2012)) & -\ ((1)/(1)-(1)/(2))+((1)/(3)-(1)/(4))+ +((1)/(2023)-(1)/(2024))\ =(1)/(2024) & ((1)/(α+1)+(1)/(α+2)+ +(1)/(α+2012)) & -\ ((1)/(1)+(1)/(2)+(1)/(3)+(1)/(4))+ +(1)/(2023). & .-(1)/(2024)-2((1)/(2)+(1)/(4)+ +(1)/(2022))\ =(1)/(2024) & aligned aligned & ((1)/(α+1)+(1)/(α+2)+ +(1)/(α+2012)) & -((1)/(1)+(1)/(2)+ +(1)/(2023)) & +(1)/(2024)+((1)/(1)+(1)/(2)+ +(1)/(1011))=(1)/(2024) & (1)/(α+1)+(1)/(α+2)+ +(1)/(α+2012) & =(1)/(1012)+(1)/(1013)+ +(1)/(2023) & α=1011 aligned

JEE Main 20249 AprilEvening Shift

Exam: JEE Main
Year: 2024
Date: 9 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Sequences and Series

JEE Main 2024 · 9 April · Evening Shift · Mathematics · Sequences and Series

If ((1)/(α+1)+(1)/(α+2)+ +(1)/(α+1012))-((1)/(2 · 1)+(1)/(4 ·...

\left(\frac{1}{\alpha+1}+\frac{1}{\alpha+2}+\ldots \ldots+\frac{1}{\alpha+1012}\right)-\left(\frac{1}{2 \cdot 1}+\frac{1}{4 \cdot 3}+\frac{1}{6 \cdot 5}+\ldots . .+\frac{1}{2024 \cdot 2023}\right)=\frac{1}{2024}

, then

\alpha

is equal to________

Correct answer

Step-by-step solution

\begin{aligned} & \left(\frac{1}{\alpha+1}+\frac{1}{\alpha+2}+\ldots+\frac{1}{\alpha+2012}\right) \\ & -\left\{\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\ldots+\left(\frac{1}{2023}-\frac{1}{2024}\right)\right\}=\frac{1}{2024} \\ & \Rightarrow\left(\frac{1}{\alpha+1}+\frac{1}{\alpha+2}+\ldots+\frac{1}{\alpha+2012}\right) \\ & -\left\{\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)+\ldots+\frac{1}{2023}\right. \\ & \left.-\frac{1}{2024}-2\left(\frac{1}{2}+\frac{1}{4}+\ldots+\frac{1}{2022}\right)\right\}=\frac{1}{2024} \\ & \end{aligned}

\begin{aligned} & \Rightarrow\left(\frac{1}{\alpha+1}+\frac{1}{\alpha+2}+\ldots+\frac{1}{\alpha+2012}\right) \\ & \quad-\left(\frac{1}{1}+\frac{1}{2}+\ldots+\frac{1}{2023}\right) \\ & \quad+\frac{1}{2024}+\left(\frac{1}{1}+\frac{1}{2}+\ldots+\frac{1}{1011}\right)=\frac{1}{2024} \\ & \Rightarrow \frac{1}{\alpha+1}+\frac{1}{\alpha+2}+\ldots+\frac{1}{\alpha+2012} \\ & \quad=\frac{1}{1012}+\frac{1}{1013}+\ldots+\frac{1}{2023} \\ & \Rightarrow \alpha=1011\end{aligned}

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