Question Number 13327 by Nayon last updated on 18/May/17
$${how}\:{can}\:{we}\:{expand}\:\left(\mathrm{1}+{x}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} ?? \\ $$
Answered by prakash jain last updated on 19/May/17
$$\left(\mathrm{1}+{x}\right)^{{y}} =\mathrm{1}+{yx}+\frac{{y}\left({y}−\mathrm{1}\right)}{\mathrm{2}!}{x}^{\mathrm{2}} +\frac{{y}\left({y}−\mathrm{1}\right)\left({y}−\mathrm{2}\right)}{\mathrm{3}!}{x}^{\mathrm{3}} + \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:…\:\:{to}\:{infinity} \\ $$$$\mathrm{The}\:\mathrm{above}\:\mathrm{series}\:\mathrm{expansion} \\ $$$$\mathrm{is}\:\mathrm{valid}\:\mathrm{for}\:\mid{x}\mid<\mathrm{1}\:\left({for}\:{convergence}\right) \\ $$$$\left(\mathrm{1}+{x}\right)^{\mathrm{1}/\mathrm{2}} =\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}{x}+\frac{\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}\right)}{\mathrm{2}!}{x}^{\mathrm{2}} +… \\ $$