Chứng minh rằng:
a) \(\small 11^{10} - 1\) chia hết cho 100;
b) \(\small 101^{100} - 1\) chia hết cho 10 000;
c) \(\small \sqrt{10}[(1+\sqrt{10})^{100}-(1-\sqrt{10})^{100}]\) là một số nguyên.
Câu a:
Ta có:
\(11^{10}- 1 = (1 + 10)^{10} =C_{10}^{0}.10^{10}+C_{10}^{1}.10^9+...\)\(+ C_{10}^{8}.10^2+C_{10}^{9}.10+C_{10}^{10}\)
\(=100(C_{10}^{0}.10^8+C_{10}^{1}.10^7+...+ C_{10}^{8}+1)+1\)
Tổng sau cùng chia hết cho 100 suy ra 1110 – 1 chia hết cho 100.
Câu b:
Ta có \(101^{100}=(100+1)^{100}=C_{100}^{0}.100^{100}\)
\(+C_{100}^{1}.100^{99}+...+ C_{100}^{99}.100+C_{100}^{100}\)
\(=100^2\left [ C_{100}^{0}.100^{98}+C_{100}^{1}.100^{97}+...+1 \right ]\)
Vậy \(101^{100}=10000\left [ C_{100}^{0}.100^{98}+C_{100}^{1}.100^{97}+...+1 \right ]\) chia hết cho 10 000.
Câu c:
Ta có \((1+\sqrt{10})^{100}=C_{100}^{0}+C_{100}^{1}\sqrt{10}+C_{100}^{2}\sqrt{10^2}+...+\)\(C_{100}^{99}\sqrt{10^{99}}+C_{100}^{100}\)
\((1-\sqrt{10})^{100}=C_{100}^{0}+C_{100}^{1}\sqrt{10}+C_{100}^{2}\sqrt{10^2}+...-\)\(C_{100}^{99}\sqrt{10^{99}}+C_{100}^{100}\)
Do đó: \((1+\sqrt{10})^{100}-(1-\sqrt{10})^{100}=2 \left ( C_{100}^{0}+C_{100}^{1}\sqrt{10}+C_{100}^{2}\sqrt{10^2}+...+ C_{100}^{99}\sqrt{10^{99}}\right )\)
Vậy nên: \(\sqrt{40}\left [ (1+\sqrt{10})^{100}-(1-\sqrt{10})^{100} \right ].\)
-- Mod Toán 11
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