Question-55055

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}

\dots .

T_{k} = \frac{k - 1}{k} \times \frac{k + 1}{k} \times \frac{k + 1}{k}

T_{2} T_{3} \dots T_{k} = \frac{1}{k} \times \frac{k + 1}{2} \times \frac{k + 1}{2} = \frac{{(k + 1)}^{2}}{4 k} > 2012

k^{2} + 2 k + 1 > 8048 k

k^{2} - 8046 k + 1 > 0

{(k - 4023)}^{2} > 4023^{2} - 1

\Rightarrow k > 4023 + \sqrt{4023^{2} - 1}

\Rightarrow k_{m i n} = 4023 + 4023 = 8046

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