Question Number 78160 by Lontum Hans last updated on 14/Jan/20
![find the term independent of x in [ ((x−1)/x)]^9](Q78160.png)
$$\mathrm{find}\:\mathrm{the}\:\mathrm{term}\:\mathrm{independent}\:\mathrm{of}\:\mathrm{x}\:\mathrm{in} \\ $$$$\:\:\left[\:\:\frac{\mathrm{x}−\mathrm{1}}{\mathrm{x}}\right]^{\mathrm{9}} \\ $$
Commented by mathmax by abdo last updated on 14/Jan/20
$$\left(\frac{{x}−\mathrm{1}}{{x}}\right)^{\mathrm{9}} =\left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right)^{\mathrm{9}} =−\left(\frac{\mathrm{1}}{{x}}−\mathrm{1}\right)^{\mathrm{9}} \:=−\sum_{{k}=\mathrm{0}} ^{\mathrm{9}} \:{C}_{\mathrm{9}} ^{{k}} \:\left(\frac{\mathrm{1}}{{x}}\right)^{{k}} \left(−\mathrm{1}\right)^{\mathrm{9}−{k}} \\ $$$$=−\sum_{{k}=\mathrm{0}} ^{\mathrm{9}} \:\frac{{C}_{\mathrm{9}} ^{{k}} }{{x}^{{k}} }\:\:\:\:{we}\:{get}\:{the}\:{independant}\:{term}\:{of}\:{x}\:{for}\:{k}=\mathrm{0}\:\Rightarrow \\ $$$${a}_{\mathrm{0}} =−{C}_{\mathrm{9}} ^{\mathrm{0}} \left(−\mathrm{1}\right)^{\mathrm{9}} \:=\mathrm{1} \\ $$