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Question Number 113593 by bemath last updated on 14/Sep/20

$$\frac{6}{5\times 14}+\frac{6}{14\times 23}+\frac{6}{23\times 32}+...+\frac{6}{(9n-4)(9n+5)}=?$$

Commented by bemath last updated on 14/Sep/20

$$thankyoubothsir$$

Answered by bobhans last updated on 14/Sep/20

$$\frac{6}{5.14}+\frac{6}{14.23}+\frac{6}{23.32}+...+\frac{6}{(9\mathrm{n}-4)(9\mathrm{n}+5)}=$$$$\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}\left(\frac{6}{(9\mathrm{k}-4)(9\mathrm{k}+5}\right)=6\times \underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}\left(\frac{1}{(9\mathrm{k}-4)(9\mathrm{k}+5)}\right)$$$$=6\times \frac{1}{9}(\frac{1}{5}-\frac{1}{9\mathrm{n}+5})$$$$=\frac{2}{3}\left(\frac{9\mathrm{n}+5-5}{5(9\mathrm{n}+5)}\right)=\frac{6\mathrm{n}}{5(9\mathrm{n}+5)}$$

Answered by Dwaipayan Shikari last updated on 14/Sep/20

$$6\underset{n=1}{\sum ^n}\frac{1}{(9n-4)(9n+5)}$$$$\frac{2}{3}\underset{n=1}{\sum ^n}\frac{1}{(9n-4)}-\frac{1}{(9n+5)}$$$$\frac{2}{3}(\frac{1}{5}-\frac{1}{9n+5})=\frac{6n}{5(9n+5)}$$$$$$

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