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chore(i18n,learn): processed translations (#45599)

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camperbot
2022-04-02 14:16:30 +05:30
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curriculum/challenges/japanese/10-coding-interview-prep/project-euler/problem-318-2011-nines.md
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@ -12,7 +12,11 @@ dashedName: problem-318-2011-nines
$\sqrt{2} + \sqrt{3}$ の偶数乗を計算すると、次のようになります。
$$\begin{align} & {(\sqrt{2} + \sqrt{3})}^2 = 9.898979485566356\ldots \\\\ & {(\sqrt{2} + \sqrt{3})}^4 = 97.98979485566356\ldots \\\\ & {(\sqrt{2} + \sqrt{3})}^6 = 969.998969071069263\ldots \\\\ & {(\sqrt{2} + \sqrt{3})}^8 = 9601.99989585502907\ldots \\\\ & {(\sqrt{2} + \sqrt{3})}^{10} = 95049.999989479221\ldots \\\\ & {(\sqrt{2} + \sqrt{3})}^{12} = 940897.9999989371855\ldots \\\\ & {(\sqrt{2} + \sqrt{3})}^{14} = 9313929.99999989263\ldots \\\\ & {(\sqrt{2} + \sqrt{3})}^{16} = 92198401.99999998915\ldots \\\\ \end{align}$$
$$\begin{align} & {(\sqrt{2} + \sqrt{3})}^2 = 9.898979485566356\ldots \\\\
& {(\sqrt{2} + \sqrt{3})}^4 = 97.98979485566356\ldots \\\\ & {(\sqrt{2} + \sqrt{3})}^6 = 969.998969071069263\ldots \\\\
& {(\sqrt{2} + \sqrt{3})}^8 = 9601.99989585502907\ldots \\\\ & {(\sqrt{2} + \sqrt{3})}^{10} = 95049.999989479221\ldots \\\\
& {(\sqrt{2} + \sqrt{3})}^{12} = 940897.9999989371855\ldots \\\\ & {(\sqrt{2} + \sqrt{3})}^{14} = 9313929.99999989263\ldots \\\\
& {(\sqrt{2} + \sqrt{3})}^{16} = 92198401.99999998915\ldots \\\\ \end{align}$$
これらの累乗の分数部を見ると、先頭で連続している 9 の個数が非減少であるように見えます。 実際に、$n$ が大きいと ${(\sqrt{2} + \sqrt{3})}^{2n}$ の小数部が 1 に近付くということを証明できます。