二项式定理（急）求证：2^（6n-3)+3^（2n-1)能被11整除.

8.不会 8.=[(8x+y)^8+(8x-y)^8][(8x+y)^8-(8x-y)^8] =[8x^8+8xy+y^8+8x^8-8xy+y^8][(8x+y+8x-y)(8x+y-8x+y)] =[8x^8+8y^8][8xy] =88x^8y+88xy^8

1)当n=1时,2^6n-3 + 3^2n-1 = 2^3 + 3^1 = 8+3 = 11,能被11整除
2)假设2^6n-3 + 3^2n-1能被11整除,如果将n换成n+1时也能被11整除,则此命题成立：
2^6（n+1）-3 + 3^2（n+1）-1
= 2^6n+6-3 + 3^2n+2-1
= 2^6n+3 + 3^2n+1
= 2^6n-3+6 + 3^2n-1+2
= 2^6 * 2^6n-3 + 3^2 * 3^2n-1
= 64 * 2^6n-3 + 9 * 3^2n-1
= (55+9) * 2^6n-3 + 9 * 3^2n-1
= 55 * 2^6n-3 + 9 * (2^6n-3 + 3^2n-1)
因55 * 2^6n-3可被11整除,而2^6n-3 + 3^2n-1也可被11整除
故证明将n换成n+1时也能被11整除,此命题成立.明白吗?