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Let $a_{n} = \frac{10^{n}}{n!}$ for $r n \geq 1 .$ Then $a_{n}$ take the greatest value when $n$ equals

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detailed solution

Correct option is D

We have anan+1=10nn!×(n+1)!10n+1=n+110Note that a1a2<1, a2a3<1, a3a4<1, a4a5<1 …a8a9<1,a9a10=1,a10a11>1,a11a12>1,…Thus, a1a11>a12∴ an is greatest if n=9 or 10.

Similar Questions

Let $S_{n} = \sum_{r = 1}^{n} r!; (n > 6) . Then, S_{n} - 7 [\frac{S_{n}}{7}]$ (where, [ . ] denotes greatest integer function) is equal to

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