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Question Number 155189 by qaz last updated on 27/Sep/21
How to extract the coefficient of  term “x^n y^m  ”  in (1+((yx)/(1−x)))(1−((yx)/(1−x)))^(−1) ?
$$\mathrm{How}\:\mathrm{to}\:\mathrm{extract}\:\mathrm{the}\:\mathrm{coefficient}\:\mathrm{of}\:\:\mathrm{term}\:“\mathrm{x}^{\mathrm{n}} \mathrm{y}^{\mathrm{m}} \:''\:\:\mathrm{in}\:\left(\mathrm{1}+\frac{\mathrm{yx}}{\mathrm{1}−\mathrm{x}}\right)\left(\mathrm{1}−\frac{\mathrm{yx}}{\mathrm{1}−\mathrm{x}}\right)^{−\mathrm{1}} ? \\ $$
Answered by mr W last updated on 26/Sep/21
=((1−x+xy)/(1−x(1+y)))  =(1−x+xy)Σ_(k=0) ^∞ x^k (1+y)^k   =(1−x+xy)Σ_(k=0) ^∞ x^k Σ_(r=0) ^k C_r ^k y^r   coef. of x^n y^m  is  C_m ^n −C_m ^(n−1) +C_(m−1) ^(n−1) =2C_(m−1) ^(n−1)   example: x^5 y^4   2C_3 ^4 =8
$$=\frac{\mathrm{1}−{x}+{xy}}{\mathrm{1}−{x}\left(\mathrm{1}+{y}\right)} \\ $$$$=\left(\mathrm{1}−{x}+{xy}\right)\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}{x}^{{k}} \left(\mathrm{1}+{y}\right)^{{k}} \\ $$$$=\left(\mathrm{1}−{x}+{xy}\right)\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}{x}^{{k}} \underset{{r}=\mathrm{0}} {\overset{{k}} {\sum}}{C}_{{r}} ^{{k}} {y}^{{r}} \\ $$$${coef}.\:{of}\:{x}^{{n}} {y}^{{m}} \:{is} \\ $$$${C}_{{m}} ^{{n}} −{C}_{{m}} ^{{n}−\mathrm{1}} +{C}_{{m}−\mathrm{1}} ^{{n}−\mathrm{1}} =\mathrm{2}{C}_{{m}−\mathrm{1}} ^{{n}−\mathrm{1}} \\ $$$${example}:\:{x}^{\mathrm{5}} {y}^{\mathrm{4}} \\ $$$$\mathrm{2}{C}_{\mathrm{3}} ^{\mathrm{4}} =\mathrm{8} \\ $$
Commented by Tawa11 last updated on 26/Sep/21
great sir
$$\mathrm{great}\:\mathrm{sir} \\ $$
Commented by qaz last updated on 27/Sep/21
thanks a lot.mr W
$$\mathrm{thanks}\:\mathrm{a}\:\mathrm{lot}.\mathrm{mr}\:\mathrm{W} \\ $$

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