find-the-term-independent-of-x-in-x-1-x-9-

Question Number 78160 by Lontum Hans last updated on 14/Jan/20

$$\mathrm{find}\:\mathrm{the}\:\mathrm{term}\:\mathrm{independent}\:\mathrm{of}\:\mathrm{x}\:\mathrm{in} \\ $$$$\:\:\left[\:\:\frac{\mathrm{x}−\mathrm{1}}{\mathrm{x}}\right]^{\mathrm{9}} \\ $$

Commented by mathmax by abdo last updated on 14/Jan/20

(((x−1)/x))^9 =(1−(1/x))^9 =−((1/x)−1)^9 =−Σ_(k=0) ^9 C_9 ^k ((1/x))^k (−1)^(9−k) =−Σ_(k=0) ^9 (C_9 ^k /x^k ) we get the independant term of x for k=0 ⇒ a_0 =−C_9 ^0 (−1)^9 =1

$$\left(\frac{{x}−\mathrm{1}}{{x}}\right)^{\mathrm{9}} =\left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right)^{\mathrm{9}} =−\left(\frac{\mathrm{1}}{{x}}−\mathrm{1}\right)^{\mathrm{9}} \:=−\sum_{{k}=\mathrm{0}} ^{\mathrm{9}} \:{C}_{\mathrm{9}} ^{{k}} \:\left(\frac{\mathrm{1}}{{x}}\right)^{{k}} \left(−\mathrm{1}\right)^{\mathrm{9}−{k}} \\ $$$$=−\sum_{{k}=\mathrm{0}} ^{\mathrm{9}} \:\frac{{C}_{\mathrm{9}} ^{{k}} }{{x}^{{k}} }\:\:\:\:{we}\:{get}\:{the}\:{independant}\:{term}\:{of}\:{x}\:{for}\:{k}=\mathrm{0}\:\Rightarrow \\ $$$${a}_{\mathrm{0}} =−{C}_{\mathrm{9}} ^{\mathrm{0}} \left(−\mathrm{1}\right)^{\mathrm{9}} \:=\mathrm{1} \\ $$

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