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  • Let  <img alt="R=(5 \sqrt{5}+11)^{2 n+1} \text { and } f=R-[R]" src="htt

Let  R=(5 \sqrt{5}+11)^{2 n+1} \text { and } f=R-[R] where [.] denotes the greatest integer function. The value of R.f is

Option: 1

4^{2 n+1}


Option: 2

4^{2 n}


Option: 3

4^{2 n-1}


Option: 4

4^{-2 n}


Answers (1)

best_answer

Since f=R-[R] \quad R=f+[R] [5 \sqrt{5}+11]^{2 n+1}=f+[R] where [R] is integer

Now letf^{\prime}=[5 \sqrt{5}-11]^{2 n+1}, 0<f^{\prime}<1

f+[R]-f^{\prime}=[5 \sqrt{5}+11]^{2 n+1}-[5 \sqrt{5}-11]^{2 n+1}=\begin{aligned} & 2\left[{ }^{2 n+1} C_1(5 \sqrt{5})^{2 n}(11)^1+{ }^{2 n+1} C_3(5 \sqrt{5})^{2 n-2}(11)^3+\ldots \ldots . .\right] \\ & =2 \cdot(\text { Integer })=2 K(K \in N)=\text { Even integer } \\ & \end{aligned}

Hence f-f^{\prime}=\text { even integer }-[R] \text {, but }-1<f-f^{\prime}<1 \text {. Therefore, } f-f^{\prime}=0 \quad \therefore f=f^{\prime}

\text { Hence R. } f=R \cdot f^{\prime}=(5 \sqrt{5}+11)^{2 n+1}(5 \sqrt{5}-11)^{2 n+1}=4^{2 n+1} \text {. }

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HARSH KANKARIA

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