Question Number 111973 by bobhans last updated on 05/Sep/20
$$\frac{\mathrm{1}}{\mathrm{5}!!}\:+\:\frac{\mathrm{1}.\mathrm{3}}{\mathrm{7}!!}\:+\:\frac{\mathrm{1}.\mathrm{3}.\mathrm{5}}{\mathrm{9}!!}\:+\:…\:+\:\frac{\mathrm{1}.\mathrm{3}.\mathrm{5}.\mathrm{7}….\mathrm{95}}{\mathrm{99}!!}\:=? \\ $$
Commented by Dwaipayan Shikari last updated on 05/Sep/20
$$\mathrm{1}.\mathrm{3}.\mathrm{5}.\mathrm{7}.\mathrm{9}.\mathrm{11}…{n}=\frac{\left(\mathrm{2}{n}\right)!}{\mathrm{2}.\mathrm{4}.\mathrm{6}.\mathrm{8}.\mathrm{10}..{n}}=\frac{\left(\mathrm{2}{n}\right)!}{\mathrm{2}^{{n}} .{n}!} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\mathrm{48}} {\sum}}\frac{\frac{\left(\mathrm{2}{n}\right)!}{\mathrm{2}^{{n}} .{n}!}}{\left(\mathrm{2}{n}+\mathrm{3}\right)!!} \\ $$$${If}\:\left(\mathrm{2}{n}+\mathrm{3}\right)!!\:\:{becomes}\:\left(\mathrm{2}{n}+\mathrm{3}\right)!\:\:{then} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\mathrm{48}} {\sum}}\frac{\left(\mathrm{2}{n}\right)!}{\mathrm{2}^{{n}} .{n}!\left(\mathrm{2}{n}+\mathrm{3}\right)\left(\mathrm{2}{n}+\mathrm{2}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}\right)!}=\underset{{n}=\mathrm{1}} {\overset{\mathrm{48}} {\sum}}\frac{\mathrm{1}}{\mathrm{2}^{{n}} .{n}!\left(\mathrm{2}{n}+\mathrm{3}\right)\left(\mathrm{2}{n}+\mathrm{2}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)} \\ $$$$ \\ $$
Commented by Dwaipayan Shikari last updated on 05/Sep/20
