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  • <img alt="\text { Let } R=(6 \sqrt{6}+14)^{2 n+1} \text { and } f=R-[R] \text {, where [.] }" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Ctext%20%7B%20Let%20%7D%20R%3D%286%20%5Csqrt%7B6

\text { Let } R=(6 \sqrt{6}+14)^{2 n+1} \text { and } f=R-[R] \text {, where [.] }  denotes the greatest integer function. The value of R f, n \in N \text { is } 

Option: 1

(25)^{2 n+1}


Option: 2

(20)^{2 n+1}


Option: 3

(16)^{2 n+1}


Option: 4

(14)^{2 n+1}


Answers (1)

best_answer

Since, f = R – [R]

\therefore R = [R] + f

\Rightarrow(6 \sqrt{6}+14)^{2 n+1}=[R]+f

where [R] is integer and 0 \leq f<1 

\text { Now, let } f^{\prime}=(6 \sqrt{6}-14)^{2 n+1}, 0<f^{\prime}<1

\begin{aligned} & \text { Also }[R]+f-f^{\prime}=(6 \sqrt{6}+14)^{2 n+1}-(6 \sqrt{6}-14)^{2 n+1} \\ & =2\left\{{ }^{2 n+1} C_1(6 \sqrt{6})^{2 n} 14+{ }^{2 n+1} C_3(6 \sqrt{6})^{2 n-2}(14)^3+\ldots\right\} \\ & =2(\text { Integer })=2 k(k \in N) \end{aligned}

= even integer

Hence f – f ′ = even integer – [R], but –1 < f – f ′ < 1
\becauseR.H.S. is integer, hence L.H.S. is also integer.
Therefore f – f ′ = 0
\therefore f = f ′

\text { Hence } R f=R f^{\prime}=(6 \sqrt{6}+14)^{2 n+1}(6 \sqrt{6}-14)^{2 n+1}=(20)^{2 n+1}

Posted by

Rakesh

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