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Question Number 212432 by hardmath last updated on 13/Oct/24

$$\frac{9}{2\cdot 4}+\frac{9}{4\cdot 6}+...+\frac{9}{4\mathrm{n}\cdot (\mathrm{n}+1)}=\frac{15}{8}$$$$\mathrm{Find}:\mathbf{n}=?$$

Answered by Ar Brandon last updated on 13/Oct/24

$$9\underset{k=1}{\sum ^n}\frac{1}{4n(n+1)}=\frac{15}{8}$$$$\underset{k=1}{\sum ^n}(\frac{1}{4n}-\frac{1}{4(n+1)})=\frac{5}{24}$$$$(\frac{1}{4}-\frac{1}{8})+(\frac{1}{8}-\frac{1}{12})+(\frac{1}{12}-\frac{1}{16})+$$$$\cdot \cdot \cdot +(\frac{1}{4(n-1)}-\frac{1}{4n})+(\frac{1}{4n}-\frac{1}{4(n+1)})=\frac{5}{24}$$$$\frac{1}{4}-\frac{1}{4(n+1)}=\frac{5}{24}\Rightarrow \frac{1}{4(n+1)}=\frac{1}{24}\Rightarrow n+1=6,n=5$$

Commented by hardmath last updated on 14/Oct/24

$$\mathrm{dear}\mathrm{professor},$$$$$$I don't quite understand this way, is there a shortcut for this?

Commented by Ar Brandon last updated on 14/Oct/24

Hello! Which line don't you understand?

Commented by hardmath last updated on 14/Oct/24

$$\mathrm{dear}\mathrm{professor},$$$$$$I do not fully understand this way, is there any other alternative way?

Commented by Ar Brandon last updated on 14/Oct/24

$$S_n=\underset{k=1}{\sum ^n}\frac{1}{4n}\Rightarrow S_{n+1}=\underset{k=1}{\sum ^n}\frac{1}{4(n+1)}=S_n-\frac{1}{4}+\frac{1}{4(n+1)}$$$$\Rightarrow \underset{k=1}{\sum ^n}(\frac{1}{4n}-\frac{1}{4(n+1)})=S_n-S_{n+1}=S_n-(S_n-\frac{1}{4}+\frac{1}{4(n+1)})$$$$\Rightarrow \underset{k=1}{\sum ^n}(\frac{1}{4n}-\frac{1}{4(n+1)})=\frac{1}{4}-\frac{1}{4(n+1)}=\frac{5}{24},n=5$$

Commented by Ar Brandon last updated on 14/Oct/24

$$\underset{k=1}{\sum ^n}(\frac{1}{4n}-\frac{1}{4(n+1)})=\frac{5}{24}$$$$\Rightarrow \frac{1}{4}-\cancel{\frac{1}{8}}$$$$+\cancel{\frac{1}{8}}-\cancel{\frac{1}{12}}$$$$+\cancel{\frac{1}{12}}-\cancel{\frac{1}{16}}$$$$+⋮$$$$+\cancel{\frac{1}{4(n-1)}}-\cancel{\frac{1}{4n}}$$$$+\cancel{\frac{1}{4n}}-\frac{1}{4(n+1)}$$$$----------$$$$\frac{1}{4}-\frac{1}{4(n+1)}=\frac{5}{24}\Rightarrow n=5$$

Commented by hardmath last updated on 14/Oct/24

$$\mathbf{T}\mathrm{hank}\mathbf{Y}\mathrm{ou}\mathbf{D}\mathrm{ear}\mathbf{P}\mathrm{rofessor}$$

Answered by BaliramKumar last updated on 15/Oct/24

$$\frac{9}{4}[\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+.....+\frac{1}{n(n+1)}]=\frac{15}{8}$$$$[\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+.....+\frac{1}{n(n+1)}]=\frac{15}{8}\times \frac{4}{9}$$$$\frac{2-1}{1\cdot 2}+\frac{3-2}{2\cdot 3}+.....+\frac{(n+1)-n}{n(n+1)}=\frac{5}{6}$$$$(\frac{1}{1}-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+.....+(\frac{1}{n}-\frac{1}{n+1})=\frac{5}{6}$$$$\frac{1}{1}-\cancel{\frac{1}{2}}+\cancel{\frac{1}{2}}-\cancel{\frac{1}{3}}+....+\cancel{\frac{1}{n}}-\frac{1}{n+1}=\frac{5}{6}$$$$\frac{1}{1}-\frac{1}{n+1}=\frac{5}{6}$$$$\frac{n}{n+1}=\frac{5}{6}$$$$n=5$$

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