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求证 对于任意正整数n,2的(6n-3)次方+3的(2n-1)次方能被11整除!

2^(6n-3)+3^(2n-1)/11=整数

2^(6n-3) + 3^(2n-1)
= 2^[3*(2n-1)] + 3^(2n-1)
= (2^3)^(2n-1) + 3^(2n-1)
= 8^(2n-1) + 3^(2n-1)

x = 8
y = 3

x^(2n-1) + y(2n-1)
= (x + y)[x^(2n-2) - x^(2n-3)*y + .... - y^(2n-2)]
= (x + y)*K
= 11K

参考证明：http://zhidao.baidu.com/question/558849.html

2^(6n-3) + 3^(2n-1)
= 2^[3*(2n-1)] + 3^(2n-1)
= (2^3)^(2n-1) + 3^(2n-1)
= 8^(2n-1) + 3^(2n-1)

x = 8
y = 3

x^(2n-1) + y(2n-1)
= (x + y)[x^(2n-2) - x^(2n-3)*y + .... - y^(2n-2)]
= (x + y)*K
= 11K

2^(6n-3) + 3^(2n-1)
= 2^[3*(2n-1)] + 3^(2n-1)
= (2^3)^(2n-1) + 3^(2n-1)
= 8^(2n-1) + 3^(2n-1)

x = 8
y = 3

x^(2n-1) + y(2n-1)
= (x + y)[x^(2n-2) - x^(2n-3)*y + .... - y^(2n-2)]
= (x + y)*K
= 11K

2^(6n-3) + 3^(2n-1)
= 2^[3*(2n-1)] + 3^(2n-1)
= (2^3)^(2n-1) + 3^(2n-1)
= 8^(2n-1) + 3^(2n-1)

x = 8
y = 3

x^(2n-1) + y(2n-1)
= (x + y)[x^(2n-2) - x^(2n-3)*y + .... - y^(2n-2)]
= (x + y)*K
= 11K
2^(6n-3) + 3^(2n-1)
= 2^[3*(2n-1)] + 3^(2n-1)
= (2^3)^(2n-1) + 3^(2n-1)
= 8^(2n-1) + 3^(2n-1)

x = 8
y = 3

x^(2n-1) + y(2n-1)
= (x + y)[x^(2n-2) - x^(2n-3)*y + .... - y^(2n-2)]
= (x + y)*K
= 11K
2^(6n-3) + 3^(2n-1)
= 2^[3*(2n-1)] + 3^(2n-1)
= (2^3)^(2n-1) + 3^(2n-1)
= 8^(2n-1) + 3^(2n-1)

x = 8
y = 3

x^(2n-1) + y(2n-1)
= (x + y)[x^(2n-2) - x^(2n-3)*y + .... - y^(2n-2)]
= (x + y)*K
= 11K

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