Question Number 106432 by Dwaipayan Shikari last updated on 05/Aug/20
Commented by Dwaipayan Shikari last updated on 05/Aug/20
https://drive.google.com/file/d/12J4x021yiK1X38OHkXoomTz_VuBlcIEt/view?usp=drivesdk
Commented by Dwaipayan Shikari last updated on 05/Aug/20
$$\left.{Respected}\:{sirs},{kindly}\:{give}\:{a}\:{try}\:{to}\:{this}\:{question}\::\right) \\ $$
Commented by Dwaipayan Shikari last updated on 05/Aug/20
$$\sqrt{\frac{\mathrm{2}}{\mathrm{3}}}=\frac{\mathrm{1}}{\:\sqrt{\frac{\mathrm{3}}{\mathrm{2}}}}=\frac{\mathrm{1}}{\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} }=\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} =\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}.\mathrm{3}}{\mathrm{2}.\mathrm{4}}.\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{1}.\mathrm{3}.\mathrm{5}}{\mathrm{2}.\mathrm{4}.\mathrm{6}}.\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{3}} +\frac{\mathrm{1}.\mathrm{3}.\mathrm{5}.\mathrm{7}}{\mathrm{2}.\mathrm{4}.\mathrm{6}.\mathrm{8}}.\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{4}} −….. \\ $$
Commented by PRITHWISH SEN 2 last updated on 05/Aug/20
$$\left(\mathrm{1}+\mathrm{x}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} =\:\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}+\frac{\mathrm{1}.\mathrm{3}}{\mathrm{2}.\mathrm{4}}\mathrm{x}^{\mathrm{2}} −\frac{\mathrm{1}.\mathrm{3}.\mathrm{5}}{\mathrm{2}.\mathrm{4}.\mathrm{6}}\mathrm{x}^{\mathrm{3}} +…. \\ $$$$\mathrm{put}\:\mathrm{x}=\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{you}\:\mathrm{will}\:\mathrm{get}\:\mathrm{the}\:\mathrm{result}. \\ $$
Commented by PRITHWISH SEN 2 last updated on 05/Aug/20
$$\mathrm{evaluate} \\ $$$$\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}.\mathrm{1}}{\mathrm{2}.\mathrm{4}}+\frac{\mathrm{1}.\mathrm{1}.\mathrm{3}}{\mathrm{2}.\mathrm{4}.\mathrm{6}}−\frac{\mathrm{1}.\mathrm{1}.\mathrm{3}.\mathrm{5}}{\mathrm{2}.\mathrm{4}.\mathrm{6}.\mathrm{8}}+… \\ $$