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Question Number 157222 by amin96 last updated on 21/Oct/21

Answered by Dimitri_01 last updated on 21/Oct/21

$$\mathrm{A}=\underset{n=1}{\sum ^{25}}\frac{1}{n(n+100)}=\frac{1}{100}\underset{n=1}{\sum ^{25}}\frac{(n+100)-n}{n(n+100)}$$$$=\frac{1}{100}\underset{n=1}{\sum ^{25}}(\frac{1}{n}-\frac{1}{n+100})=\frac{1}{100}(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+\frac{1}{3}-\frac{1}{103}+\cdot \cdot \cdot +\frac{1}{25}-\frac{1}{125})$$$$$$

Answered by qaz last updated on 21/Oct/21

$$\frac{\mathrm{A}}{\mathrm{B}}=\frac{\underset{\mathrm{n}=1}{\sum ^{25}}\frac{1}{\mathrm{n}(100+\mathrm{n})}}{\underset{\mathrm{n}=1}{\sum ^{100}}\frac{1}{\mathrm{n}(\mathrm{n}+25)}}$$$$=\frac{\underset{\mathrm{n}=1}{\sum ^{25}}(\frac{1}{\mathrm{n}}-\frac{1}{\mathrm{n}+100})}{4\underset{\mathrm{n}=1}{\sum ^{100}}(\frac{1}{\mathrm{n}}-\frac{1}{\mathrm{n}+25})}$$$$=\frac{{\mathrm{H}}_{25}-{\mathrm{H}}_{125}+{\mathrm{H}}_{100}}{4({\mathrm{H}}_{100}-{\mathrm{H}}_{125}+{\mathrm{H}}_{25})}$$$$=\frac{1}{4}$$

Commented by Tawa11 last updated on 21/Oct/21

$$\mathrm{Great}\mathrm{sirs}$$

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