（急!证明对于任意自然数n,3^(n+2) - 2^(n+3)+3^n-2^(n+1)一定能被10整

（急!证明对于任意自然数n,3^(n+2) - 2^(n+3)+3^n-2^(n+1)一定能被10整

3^(n+2) - 2^(n+3) + 3^n - 2^(n+1) =(3^2*3^n+3^n)-(2^3*2^n+2*2^n) =3^n(3^2+1)-2^n(2^3+2) =10*3^n-10*2^n =10*(3^n-2^n) 所以一定能被10整除 a-b=2,a-c=0.5 相减 (a-b)-(a-c)=2-0.5 c-b=1.5 b-c=-1.5 （b-c）^2 - 3(b-c)+9／4 =2.25+4.5+2.25 =9 (3x-y-2)(2x-y-1). =6x^2-3xy-3x-2xy+y^2+y-4x+2y+2 =6x^2-5xy+y^2-7x+3y+2 =6x^2-5xy+y^2+mx+ny+2 对应项系数相等 所以m=-7,n=3======以下答案可供参考======供参考答案1:(1)3^(n+2) - 2^(n+3) + 3^n - 2^(n+1) =3^(n+2)+3^n-(2^(n+3)+2^(n+1))=3^n(1+3^2)-2^(n+1)(1+2^2)=3^n*10-2^(n+1)*5=10(3^n-2^n)所以原式一定能被10整除(2)b-c=0.5-2=-1.5原式=（b-c-3/2）^2=(-1.5-1.5)^2=9(3)将右边化开，得(3x-y-2)(2x-y-1)=6x^2-5xy+y^2-7x+3y+2与左边对应，立得m=-7,n=3供参考答案2:（1）原式=10（3^n-2^n）(2)原式=(b-c-3/2)^2=9(3)右边=6x^2-5xy+y^2-7x+3y+2供参考答案3:(1)(n+2) - 2^(n+3) + 3^n - 2^(n+1) =3^n*9-2^n*8+3^n-2^n+2 =3^n(9+1)-2^n(8+2) =10(3^n-2^n)(2)(a-c)-(a-b)=b-c=-1.5（b-c）^2 - 3(b-c)+9／4=(b-c-3/2)^2=0(3)已知6x^2-5xy+y^2+mx+ny+2=(3x-y-2)(2x-y-1) (3x-y-2)(2x-y-1) =6x^2-5xy+y^2-7x+3y+2 m=-7 ,n=3

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