$\sqrt{2}, \sqrt{\boxed{16}}, \boxed{17}, \sqrt{\boxed{18}}, 3\sqrt{6}, ...$ が等比数列になるように、空欄を埋める問題です。

1. 問題の内容

$\sqrt{2}, \sqrt{\boxed{16}}, \boxed{17}, \sqrt{\boxed{18}}, 3\sqrt{6}, ...$ が等比数列になるように、空欄を埋める問題です。

2. 解き方の手順

まず、この等比数列の公比 $r$ を求めます。初項は $a = \sqrt{2}$ であり、第5項は $3\sqrt{6}$ です。等比数列の一般項は $a_n = ar^{n-1}$ で表されるので、第5項は $a_5 = ar^4$ となります。したがって、

$3\sqrt{6} = \sqrt{2} r^4$

$r^4 = \frac{3\sqrt{6}}{\sqrt{2}} = 3\sqrt{3}$

$r = \sqrt[4]{3\sqrt{3}} = \sqrt[4]{3^{\frac{3}{2}}} = 3^{\frac{3}{8}}$

となります。

第2項は $ar = \sqrt{2} \cdot 3^{\frac{3}{8}} = \sqrt{2 \cdot 3^{\frac{3}{4}}} = \sqrt{2 \sqrt[4]{27}} = \sqrt{\sqrt[4]{16} \sqrt[4]{27}} = \sqrt{\sqrt[4]{432}} = \sqrt[8]{432}$

より $\sqrt{\boxed{16}}$ は $\sqrt[8]{432}$ なので、$16$ に入る値は$432^{1/4} \approx 4.5546$

第3項は $ar^2 = \sqrt{2} \cdot (3^{\frac{3}{8}})^2 = \sqrt{2} \cdot 3^{\frac{3}{4}} = \sqrt{2 \cdot 3^{\frac{3}{2}}} = \sqrt{2 \cdot 3\sqrt{3}} = \sqrt{6\sqrt{3}} = \sqrt[4]{36\cdot 3} = \sqrt[4]{108}$

より $\boxed{17}$ に入る値は$\sqrt[4]{108} \approx 3.226$

第4項は $ar^3 = \sqrt{2} \cdot (3^{\frac{3}{8}})^3 = \sqrt{2} \cdot 3^{\frac{9}{8}} = \sqrt{2 \cdot 3^{\frac{9}{4}}} = \sqrt{2 \cdot 3^2 \sqrt[4]{3^2}} = \sqrt{18\sqrt{3}} = \sqrt{\sqrt{18^2}\sqrt{3}} = \sqrt{\sqrt{324\cdot 3}} = \sqrt{\sqrt{972}} = \sqrt[4]{972}$

より $\sqrt{\boxed{18}}$ は $\sqrt[4]{972}$ なので、$18$ に入る値は $\sqrt{972} \approx 31.1769$

より正確な値を計算します。

$r^4 = \frac{3\sqrt{6}}{\sqrt{2}} = 3\sqrt{3}$

$a_2 = \sqrt{2}r = \sqrt{\boxed{16}}$

$a_2^2 = 2r^2$

$r^2 = \sqrt{3\sqrt{3}} = 3^{3/4}$

$a_2^2 = 2 \cdot 3^{3/4}$

$\sqrt{16} = a_2 = \sqrt{2 \cdot 3^{3/4}}$

$16 = 2 \cdot 3^{3/4}$ はありえないので、$a_2$は$\sqrt{\boxed{16}}$ではなく、2乗して$\boxed{16}$になると考える

$a_2^2 = (\sqrt{2}r)^2 = 2r^2$

$a_5/a_1 = r^4 = 3\sqrt{3}$

$r = (3\sqrt{3})^{1/4}$

$a_2^2 = 2r^2 = 2\sqrt{3\sqrt{3}} = 2\cdot 3^{3/4}$

$a_3 = a_1r^2 = \sqrt{2}3^{3/4}$

$a_4 = a_1r^3 = \sqrt{2} 3^{9/8}$

$a_4^2 = 2\cdot 3^{9/4}$

$a_2^2 = 2(3\sqrt{3})^{1/2} = 2\sqrt{3\sqrt{3}}$

$a_3 = \sqrt{2}(3\sqrt{3})^{1/2} = \sqrt{2\sqrt{27}}$

$a_4^2 = ( \sqrt{2}r^3)^2 = 2r^6 = 2(3\sqrt{3})^{3/2} = 2 \times 3^{3/2+\frac{3}{4}}= 2(3)^{9/4}=2(3^2)(3^{\frac{1}{4}})=18\sqrt[4]{3}$

$r^4 = 3\sqrt{3} = 3^{3/2}$

$r = 3^{3/8}$

$a_1 = \sqrt{2} = 2^{\frac{1}{2}}$

$a_2 = \sqrt{2}(3^{\frac{3}{8}}) = \sqrt{2\cdot3^{\frac{3}{4}}}$ so 16に$\sqrt{2\cdot 3^{3/4}}$ を代入. $\sqrt[4]{2^2\cdot 3^{3}} = \sqrt[4]{4 \times 27}= \sqrt[4]{108}=3.22$

$a_3 = \sqrt{2}r^2 = 2^{1/2}(3^{3/8})^2 = \sqrt{2}(3^{6/8})=\sqrt{2}(3^{3/4})=2^{1/2}3^{3/4}$. $17に\sqrt{2}\cdot 3^{\frac{3}{4}}$を代入. $\sqrt{2} \times \sqrt[4]{27}$.

$a_4 = \sqrt{2} r^3 = \sqrt{2} (3^{3/8})^3 = 2^{1/2} 3^{9/8}$.

so $\sqrt{18}=a_4$

$18=(a_4)^2=a_4*a_4=(2^{1/2})^2(3^{9/8})^2=2(3^{18/8})=2(3^{9/4})=2*3^2 \cdot 3^{1/4}=18 \sqrt[4]{3} = 18 \cdot 1.316=23.684$.

$\sqrt{2}3^{3/8}= \sqrt{\boxed{16}}$

$16 = 2*3^{3/4}$

$16 = 2*2.2795 \approx 4.55$.

$\sqrt{17}= 2^{1/2}3^{6/8}=2^{1/2}3^{3/4}$

$17 = \sqrt{2}*2.27 = 3.20$

$\sqrt{\boxed{18}} = \sqrt{2}3^{9/8} = \sqrt[4]{2^2}3^{9/8}=18 \sqrt[4]{3}$. so $\sqrt{2}*(18* \sqrt[4]{3.0})=$

3. 最終的な答え

16：$2 \cdot 3^{\frac{3}{4}}$

17：$\sqrt{2} \cdot 3^{\frac{3}{4}}$

18：$18 \sqrt[4]{3}$