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Question Number 63860 by gunawan last updated on 10/Jul/19

$$\mathrm{If}n\mathrm{is}\mathrm{even}\mathrm{positive}\mathrm{integer},\mathrm{then}\mathrm{the}$$$$\mathrm{condition}\mathrm{that}\mathrm{the}\mathrm{greatest}\mathrm{term}\mathrm{in}\mathrm{the}$$$$\mathrm{expansion}\mathrm{of}{(1+x)}^n\mathrm{may}\mathrm{have}\mathrm{the}$$$$\mathrm{greatest}\mathrm{coefficient}\mathrm{also}\mathrm{is}$$

Answered by mr W last updated on 10/Jul/19

$$T_r=C_r^n{x}^r$$$$n=2m$$$$T_{m-1}=C_{m-1}^{2m}{x}^{m-1}$$$$T_m=C_m^{2m}{x}^{m}shouldbemaximum.$$$$T_{m+1}=C_{m+1}^{2m}{x}^{m+1}$$$$\frac{T_m}{T_{m-1}}=\frac{C_m^{2m}}{C_{m-1}^{2m}}x>1$$$$\frac{\left(2m\right)!(2m-m+1)!(m-1)!}{(2m-m)!m!\left(2m\right)!}x>1$$$$\frac{(m+1)}{m}x>1$$$$\frac{(n+2)x}{n}>1$$$$\Rightarrow x>\frac{n}{n+2}=1-\frac{2}{n+2}$$$$\frac{T_m}{T_{m+1}}=\frac{C_m^{2m}}{{xC}_{m+1}^{2m}}>1$$$$\frac{n+2}{nx}>1$$$$\frac{1}{x}>\frac{n}{n+2}$$$$\Rightarrow x<\frac{n+2}{n}=1+\frac{2}{n}$$$$$$$$\Rightarrow conditionis$$$$1-\frac{2}{n+2}<x<1+\frac{2}{n}$$

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