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Question Number 8107 by Nadium last updated on 30/Sep/16
The general term in the expansion of  (1−2x)^(3/4)   is
$$\mathrm{The}\:\mathrm{general}\:\mathrm{term}\:\mathrm{in}\:\mathrm{the}\:\mathrm{expansion}\:\mathrm{of} \\ $$$$\left(\mathrm{1}−\mathrm{2}{x}\right)^{\mathrm{3}/\mathrm{4}} \:\:\mathrm{is} \\ $$
Answered by prakash jain last updated on 30/Sep/16
If ∣x∣<1  (1+x)^y =1+xy+((y(y−1))/(2!))x^2 +...  If ∣x∣<(1/2)  (1−2x)^(3/4) =Σ_(i=0) ^∞ (((3/4)((3/4)−1)..((3/4)−(i−1)))/(i!))(2x)^i
$$\mathrm{If}\:\mid{x}\mid<\mathrm{1} \\ $$$$\left(\mathrm{1}+{x}\right)^{{y}} =\mathrm{1}+{xy}+\frac{{y}\left({y}−\mathrm{1}\right)}{\mathrm{2}!}{x}^{\mathrm{2}} +… \\ $$$$\mathrm{If}\:\mid{x}\mid<\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\left(\mathrm{1}−\mathrm{2}{x}\right)^{\mathrm{3}/\mathrm{4}} =\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\frac{\mathrm{3}}{\mathrm{4}}\left(\frac{\mathrm{3}}{\mathrm{4}}−\mathrm{1}\right)..\left(\frac{\mathrm{3}}{\mathrm{4}}−\left({i}−\mathrm{1}\right)\right)}{{i}!}\left(\mathrm{2}{x}\right)^{{i}} \\ $$

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