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Question Number 28532 by abdo imad last updated on 26/Jan/18

$$letgive{A}_n={(C_n^0.C_n^1....C_n^n)}^{\frac{1}{n+1}}find{}^n\sqrt{A}{}_n.$$

Commented by abdo imad last updated on 27/Jan/18

$$Qfind{lim}_{n\to +\mathrm{\infty }}^{}{}^n\sqrt{A_n}.$$

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