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Question Number 55055 by peter frank last updated on 16/Feb/19

Answered by mr W last updated on 16/Feb/19

$$T_n=(1+\frac{1}{n})(1-\frac{1}{n^2})=\frac{(n-1)}{n}\times \frac{(n+1)}{n}\times \frac{(n+1)}{n}$$$$T_2=\frac{1}{2}\times \frac{3}{2}\times \frac{3}{2}$$$$T_3=\frac{2}{3}\times \frac{4}{3}\times \frac{4}{3}$$$$T_4=\frac{3}{4}\times \frac{5}{4}\times \frac{5}{4}$$$$....$$$$T_k=\frac{k-1}{k}\times \frac{k+1}{k}\times \frac{k+1}{k}$$$$T_2{T}_3...T_k=\frac{1}{k}\times \frac{k+1}{2}\times \frac{k+1}{2}=\frac{{(k+1)}^2}{4k}>2012$$$$k^2+2k+1>8048k$$$$k^2-8046k+1>0$$$${(k-4023)}^2>{4023}^2-1$$$$\Rightarrow k>4023+\sqrt{{4023}^2-1}$$$$\Rightarrow k_{min}=4023+4023=8046$$

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