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Question Number 26554 by das47955@mail.com last updated on 26/Dec/17

$$\left(2\right)\mathbf{Find}\mathbf{the}\mathbf{middle}\mathbf{trem}\left(\mathbf{s}\right)\mathbf{in}\mathbf{the}$$$$\mathbf{expansion}\mathbf{of}\mathbf{following}-$$$${({\mathbf{x}}^2+\frac{1}{{\mathbf{x}}^3})}^{14}$$

Answered by prakash jain last updated on 26/Dec/17

$$\mathrm{total}\mathrm{number}\mathrm{of}\mathrm{terms}n+1=15$$$$\mathrm{middle}\mathrm{term}=8^{\mathrm{th}}$$$${(x^2+\frac{1}{x^3})}^{15}=\underset{i=0}{\sum ^{14}}{}^{14}{C}_i{\left(x^2\right)}^i{\left(\frac{1}{x^3}\right)}^{14-i}$$$$\mathrm{middlw}\mathrm{term}={}^{14}{C}_7{\left(x^2\right)}^7{\left(\frac{1}{x^3}\right)}^7={}^{14}{C}_7{x}^{-7}$$

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