Question Number 26554 by das47955@mail.com last updated on 26/Dec/17
$$\left(\mathrm{2}\right)\:\boldsymbol{\mathrm{Find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{middle}}\:\boldsymbol{\mathrm{trem}}\left(\boldsymbol{\mathrm{s}}\right)\:\boldsymbol{\mathrm{in}}\:\boldsymbol{\mathrm{the}} \\ $$$$\boldsymbol{\mathrm{expansion}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{following}}− \\ $$$$\:\:\:\:\:\left(\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}^{\mathrm{3}} }\right)^{\mathrm{14}} \\ $$
Answered by prakash jain last updated on 26/Dec/17
$$\mathrm{total}\:\mathrm{number}\:\mathrm{of}\:\mathrm{terms}\:{n}+\mathrm{1}=\mathrm{15} \\ $$$$\mathrm{middle}\:\mathrm{term}=\mathrm{8}^{\mathrm{th}} \\ $$$$\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\right)^{\mathrm{15}} =\underset{{i}=\mathrm{0}} {\overset{\mathrm{14}} {\sum}}\:^{\mathrm{14}} {C}_{{i}} \left({x}^{\mathrm{2}} \right)^{{i}} \left(\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\right)^{\mathrm{14}−{i}} \\ $$$$\mathrm{middlw}\:\mathrm{term}=\:\:^{\mathrm{14}} {C}_{\mathrm{7}} \left({x}^{\mathrm{2}} \right)^{\mathrm{7}} \left(\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\right)^{\mathrm{7}} =\:^{\mathrm{14}} {C}_{\mathrm{7}} {x}^{−\mathrm{7}} \\ $$