Limit(1)

\mathbf{\lim_{n\to\infty}\left\{\left(\sqrt{3}+1\right)^{2n}\right\}}=$\\\\ \textbf{Where $\mathbf{\left\{.\right\} = }$Fractional part of $\mathbf{x}$}

2 Answers

21
Shubhodip ·

Let (\sqrt{3}+1)^{2n}= I + f

where I is an integer and 0≤f<1

We have (\sqrt{3}-1)^{2n}= f_{1}

where f1<1

we have
(\sqrt{3}+1)^{2n}+ (\sqrt{3}-1)^{2n}= 2[(\sqrt{3})^{2n} + \begin{pmatrix} 2n\\ 2 \end{pmatrix}(\sqrt{3})^{2n-2}+....+1 ]

= an even integer

so I + f + f1 is an even integer \Leftrightarrow f + f_{1} is an integer

Since 0<f+ f1<2 we must have f + f1 = 1

So, f = 1- f1

But,\lim_{n\rightarrow \propto } (\sqrt{3}-1)^{2n} = 0 = f_{1}

So, f = 1- f_{1} = 1
\lim_{n\rightarrow \propto }\left\{(\sqrt{3}+1)^{2n} \right\} = 1

1708
man111 singh ·

Yes Shubhodip you are Right Ans = 1

Thanks.

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