Question Number 113593 by bemath last updated on 14/Sep/20
$$\frac{\mathrm{6}}{\mathrm{5}×\mathrm{14}}+\frac{\mathrm{6}}{\mathrm{14}×\mathrm{23}}+\frac{\mathrm{6}}{\mathrm{23}×\mathrm{32}}+...+\frac{\mathrm{6}}{\left(\mathrm{9}{n}−\mathrm{4}\right)\left(\mathrm{9}{n}+\mathrm{5}\right)}=? \\ $$
Commented by bemath last updated on 14/Sep/20
$${thank}\:{you}\:{both}\:{sir} \\ $$
Answered by bobhans last updated on 14/Sep/20
$$\frac{\mathrm{6}}{\mathrm{5}.\mathrm{14}}+\frac{\mathrm{6}}{\mathrm{14}.\mathrm{23}}+\frac{\mathrm{6}}{\mathrm{23}.\mathrm{32}}+...+\frac{\mathrm{6}}{\left(\mathrm{9n}−\mathrm{4}\right)\left(\mathrm{9n}+\mathrm{5}\right)}\:= \\ $$$$\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\left(\frac{\mathrm{6}}{\left(\mathrm{9k}−\mathrm{4}\right)\left(\mathrm{9k}+\mathrm{5}\right.}\right)\:=\:\mathrm{6}\:×\:\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\left(\frac{\mathrm{1}}{\left(\mathrm{9k}−\mathrm{4}\right)\left(\mathrm{9k}+\mathrm{5}\right)}\right) \\ $$$$=\:\mathrm{6}×\frac{\mathrm{1}}{\mathrm{9}}\left(\frac{\mathrm{1}}{\mathrm{5}}−\frac{\mathrm{1}}{\mathrm{9n}+\mathrm{5}}\right) \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}\left(\frac{\mathrm{9n}+\mathrm{5}−\mathrm{5}}{\mathrm{5}\left(\mathrm{9n}+\mathrm{5}\right)}\right)\:=\:\frac{\mathrm{6n}}{\mathrm{5}\left(\mathrm{9n}+\mathrm{5}\right)} \\ $$
Answered by Dwaipayan Shikari last updated on 14/Sep/20
$$\mathrm{6}\underset{{n}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{9}{n}−\mathrm{4}\right)\left(\mathrm{9}{n}+\mathrm{5}\right)} \\ $$$$\frac{\mathrm{2}}{\mathrm{3}}\underset{{n}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{9}{n}−\mathrm{4}\right)}−\frac{\mathrm{1}}{\left(\mathrm{9}{n}+\mathrm{5}\right)} \\ $$$$\frac{\mathrm{2}}{\mathrm{3}}\left(\frac{\mathrm{1}}{\mathrm{5}}−\frac{\mathrm{1}}{\mathrm{9}{n}+\mathrm{5}}\right)=\frac{\mathrm{6}{n}}{\mathrm{5}\left(\mathrm{9}{n}+\mathrm{5}\right)} \\ $$$$ \\ $$