Find-the-value-of-3-17-2-6-

Question Number 118876 by bemath last updated on 20/Oct/20

F i n d t h e v a l u e o f ⌊ {(\frac{3 + \sqrt{17}}{2})}^{6} ⌋ .

Answered by bobhans last updated on 20/Oct/20

2040

Answered by MJS_new last updated on 20/Oct/20

{(\frac{3 + \sqrt{17}}{2})}^{6} = \frac{2041 + 495 \sqrt{17}}{2}

0 ⩽ \frac{2041}{2} + \frac{495 \sqrt{17}}{2} - n < 1 with n \in N

2 n - 2041 ⩽ 495 \sqrt{17} < 2 n - 2039

all are positive \Rightarrow squaring allowed

{(2 n - 2041)}^{2} ⩽ 4165435 < {(2 n - 2039)}^{2}

\Rightarrow

n^{2} - 2041 n + 64 ⩽ 0 < n^{2} - 2039 n - 1976

n^{2} - 2041 n + 64 ⩽ 0 \Rightarrow \frac{2041 - 495 \sqrt{17}}{2} ⩽ n ⩽ \frac{2041 + 495 \sqrt{17}}{2}

n^{2} - 2039 n - 1976 > 0 \Rightarrow n < \frac{2039 - 495 \sqrt{17}}{2} \lor n > \frac{2039 + 495 \sqrt{17}}{2}

\Rightarrow \frac{2039 + 495 \sqrt{17}}{2} < n ⩽ \frac{2041 + 495 \sqrt{17}}{2}

2039.96 \dots < n ⩽ 2040.96 \dots

but n \in N

\Rightarrow n = 2040

Answered by TANMAY PANACEA last updated on 20/Oct/20

y = \sqrt{x}

\frac{d y}{d x} \approx \frac{△ y}{△ x} = \frac{1}{2 \sqrt{x}} \to △ y = \frac{△ x}{2 \sqrt{x}}

w h e n x = 16 y = \sqrt{16} = 4

x + △ x = 17 y + △ y = ?

△ y = \frac{1}{2 \sqrt{16}} = \frac{1}{8}

\sqrt{17} = 4 + \frac{1}{8} = \frac{33}{8}

{(\frac{3 + \frac{33}{8}}{2})}^{6} = {(\frac{3 + 4.125}{2})}^{6} = {(1.5 + 2.0625)}^{6} = {(3.5625)}^{6}