Question

Chứng minh rằng :

a) \(11^{10}-1\) chia hết cho 100

b) \(101^{100}-1\) chia hết cho 10 000

c) \(\sqrt{10}\left[\left(1+\sqrt{10}\right)^{100}-\left(1-\sqrt{10}\right)^{100}\right]\) là một số nguyên 

Answer 1

a) 1110 – 1 = (1 + 10)10 – 1 = (1 + C110 10 + C210102 + … +C910 109 + 1010) – 1

= 102 + C210102 +…+ C910 109 + 1010.

Tổng sau cùng chia hết cho 100 suy ra 1110 – 1 chia hết cho 100.

b) Ta có

101100 – 1 = (1 + 100)100 - 1

= (1 + C1100 100 + C2100 1002 + …+C99100 10099 + 100100) – 1.

= 1002 + C21001002 + …+ 10099 + 100100.

Tổng sau cùng chia hết cho 10 000 suy ra 101100 – 1 chia hết cho 10 000.

c) (1 + √10)100 = 1 + C1100 √10 + C2100 (√10)2 +…+ (√10)99 + (√10)100

(1 - √10)100 = 1 - C1100 √10 + C2100 (√10)2 -…- (√10)99 + (√10)100

√10[(1 + √10)100 – (1 - √10)100] = 2√10[C1100 √10 + C3100 (√10)3 +…+ . (√10)99]

= 2(C1100 10 + C3100 102 +…+ 1050)

Tổng sau cùng là một số nguyên, suy ra √10[(1 + √10)100 – (1 - √10)100] là một số nguyên.

Answer 2

a) \(11^{10}-1=\left(10+1\right)^{10}-1\)\(=C^0_{10}10^{10}+C^1_{10}10^9+...+C^9_{10}10+C^{10}_{10}-1\)
\(=10^{10}+C^1_{10}10^9+...+C^8_{10}10^2+10.10\) chia hết cho 100.
b) \(\left(101\right)^{100}-1=\left(100+1\right)^{100}-1\)
\(=100^{100}+C_{100}^{99}100^{99}+....+C^1_{100}100+C_{100}^{100}100^0-1\)
\(=100^{100}+C_{100}^{99}100^{99}+....+C^2_{100}100^2+100.100+1-1\)
\(=100^{100}+C_{100}^{99}100^{99}+....+C^2_{100}100^2+10000\) chia hết cho 10000.

Answer 3

c) \(\sqrt{10}\left[\left(1+\sqrt{10}\right)^{100}-\left(1-\sqrt{10}\right)^{100}\right]\)
Ta có: \(\left(1+\sqrt{10}\right)^{100}=C^0_{100}\sqrt{10}^0+C^1_{100}\sqrt{10}^1+...+C_{100}^{100}\sqrt{10}^{100}\)
\(\left(1-\sqrt{10}\right)^{100}=C^0_{100}\sqrt{10}^0-C^1_{100}\sqrt{10}^1+...+C_{100}^{100}\sqrt{10}^{100}\)
Vì vậy
\(\left(1+\sqrt{10}\right)^{100}-\left(1-\sqrt{10}\right)^{100}\)\(=2\left(C^1_{100}\sqrt{10}^1+C^3_{100}\sqrt{10}^3+...+C^{99}_{100}\sqrt{10}^{99}\right)\).
Ta có:
\(\sqrt{10}\left[\left(1+\sqrt{10}\right)^{100}-\left(1-\sqrt{10}\right)^{100}\right]\)\(=2.\sqrt{10}\left(C^1_{100}\sqrt{10}^1+C^3_{100}\sqrt{10}^3+...+C^{99}_{100}\sqrt{10}^{99}\right)\)
\(=2\left(C^1_{100}\sqrt{10}^2+C^3_{100}\sqrt{10}^4+....+C^{99}_{100}\sqrt{10}^{100}\right)\)
\(=2\left(C^1_{100}10+C^3_{100}10^2+....+C^{99}_{100}10^{50}\right)\)\(\in N\).
nên \(\sqrt{10}\left[\left(1+\sqrt{10}\right)^{100}-\left(1-\sqrt{10}\right)^{100}\right]\) là một số nguyên.