Question and Answers Forum

All Questions      Topic List

UNKNOWN Questions

Previous in All Question      Next in All Question      

Previous in UNKNOWN      Next in UNKNOWN      

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

$$\mathrm{If}\:\mid{x}\mid<\mathrm{1} \\ $$$$\left(\mathrm{1}+{x}\right)^{{y}} =\mathrm{1}+{xy}+\frac{{y}\left({y}−\mathrm{1}\right)}{\mathrm{2}!}{x}^{\mathrm{2}} +... \\ $$$$\mathrm{If}\:\mid{x}\mid<\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\left(\mathrm{1}−\mathrm{2}{x}\right)^{\mathrm{3}/\mathrm{4}} =\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\frac{\mathrm{3}}{\mathrm{4}}\left(\frac{\mathrm{3}}{\mathrm{4}}−\mathrm{1}\right)..\left(\frac{\mathrm{3}}{\mathrm{4}}−\left({i}−\mathrm{1}\right)\right)}{{i}!}\left(\mathrm{2}{x}\right)^{{i}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com