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Question Number 207661 by efronzo1 last updated on 22/May/24

$$\mathrm{P}=1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...+\frac{1}{2023}$$$$\mathrm{Q}=\frac{1}{1\times 2023}+\frac{1}{3\times 2021}+\frac{1}{5\times 2019}+...+\frac{1}{2023\times 1}$$$$\frac{\mathrm{P}}{\mathrm{Q}}=?$$

Answered by Frix last updated on 22/May/24

$$P\left(m\right)=\underset{k=1}{\sum ^{\frac{m+1}{2}}}\frac{1}{2k-1}\wedge m=2n-1$$$$P(2n-1)=\underset{k=1}{\sum ^n}\frac{1}{2k-1}=H_{2n-1}-\frac{H_{n-1}}{2}$$$$Q\left(m\right)=\underset{k=1}{\sum ^{\frac{m+1}{2}}}\frac{1}{(2k-1)(m+1-(2k-1)}\wedge m=2n-1$$$$Q(2n-1)=\underset{k=1}{\sum ^n}\frac{1}{(2k-1)(2n-2k+1)}=$$$$=\frac{1}{2n}(\underset{k=1}{\sum ^n}\frac{1}{2k-1}+\underset{k=1}{\sum ^n}\frac{1}{2n-2k+1})=$$$$[\underset{k=1}{\sum ^n}\frac{1}{2k-1}=\underset{k=1}{\sum ^n}\frac{1}{2n-2k+1}]$$$$=\frac{1}{n}(H_{2n-1}-\frac{H_{n-1}}{2})$$$$\Rightarrow$$$$\frac{P(2n-1)}{Q(2n-1)}=n$$$$\Rightarrow$$$$\frac{P\left(m\right)}{Q\left(m\right)}=\frac{m+1}{2}$$$$m=2023\Rightarrow \mathrm{Answer}\mathrm{is}1012$$

Answered by MM42 last updated on 22/May/24

$$Q=\frac{1}{2024}[(1+\frac{1}{2023})+(\frac{1}{3}+\frac{1}{2021})+(\frac{1}{5}+\frac{1}{2019})+...+(\frac{1}{2019}+\frac{1}{5})+(\frac{1}{2021}+\frac{1}{3})+(\frac{1}{2023}+1)]$$$$=\frac{1}{2024}\left[2p\right]=\frac{p}{1012}$$$$\Rightarrow \frac{P}{Q}=1012✓$$$$$$

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