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

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