If ((1)/( +1)+(1)/( +2)+ ..+(1)/( +1012))-((1)/(2 1)+(1)/(4 3)+(1)/(6 5)+ +(1)/(2024 2023)…

Question

If $(\frac{1}{\alpha +1}+\frac{1}{\alpha +2}+\ldots ..+\frac{1}{\alpha +1012})-(\frac{1}{2\cdot 1}+\frac{1}{4\cdot 3}+\frac{1}{6\cdot 5}+\ldots \ldots +\frac{1}{2024\cdot 2023})=\frac{1}{2024}$, then $\alpha $ is equal to \_\_\_\_\_\_\_\_\_\_\_.

Answer: 1011

Accepted Answer

$\begin{gathered} \frac{1}{\alpha +1}+\frac{1}{\alpha +2}+\ldots +\frac{1}{\alpha +2012}- \ (\frac{1}{2	imes 21}+\frac{1}{4	imes 3}+\ldots +\frac{1}{2024}\cdot \frac{1}{2023})=\frac{1}{2024} \ \, \sum _{r=1}^{1012}\frac{1}{2r(2r-1)}=\sum _{r=1}^{1012}(\frac{1}{2r-1}-\frac{1}{2r}) \ \, =(1-\frac{1}{2})+(\frac{1}{3}-\frac{1}{4})+\ldots +(\frac{1}{2023}-\frac{1}{2024}) \ \, =(1+\frac{1}{3}+\frac{1}{5}+\ldots .+\frac{1}{2023}) \ \, -(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\ldots +\frac{1}{2024}) \ \, =(1+\frac{1}{3}+\ldots +\frac{1}{2023})-\frac{1}{2}(1+\frac{1}{2}+\frac{1}{3}+\ldots +\frac{1}{1012}) \end{gathered}$

$\begin{gathered} =(1+\frac{1}{2}+\frac{1}{3}+\ldots +\frac{1}{2023})-\frac{1}{2}(1+\frac{1}{2}+\ldots +\frac{1}{1011}) \ \frac{-1}{2}(1+\frac{1}{2}+\ldots +\frac{1}{1012}) \ =\frac{1}{1012}+\frac{1}{1013}+\ldots +\frac{1}{2023}-\frac{1}{2024} \ \Rightarrow \alpha +1012=2023 \ \Rightarrow \alpha =1011 \end{gathered}$

Solutions · 1