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If N=√5328/96...

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If N=√5328/961, then (31×N) is

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\sqrt{3^{n}} = 81

n = ?

\sqrt{3^{n}} = 81

. Then n is equal to

lim n \to \infty \frac{n . 3^{n}}{n {(x - 2)}^{n} + n . 3^{n + 1} - 3^{n}} = \frac{1}{3}

, then the range of

x

n \in N

)

^{3 n} P_{n - 1} :^{3 n + 1} P_{n} = 1 : 4,

n

\sqrt{\sqrt{2500} + \sqrt{961}} = {(x)}^{2}

x

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