求证数学归纳法,假设(3^(6k+3)+4^(6k+3))/13是整数证(3^(6(k+1)+3)+4^(6(k+1)+3))/13是整数

Question

求证数学归纳法,假设(3^(6k+3) + 4^(6k+3))/13 是整数
证(3^(6(k+1)+3) + 4^(6(k+1)+3))/13 是整数

Answer 1

(3^(6(k+1)+3) + 4^(6(k+1)+3))/13
=[3^(6k+3+6) + 4^(6k+3+6)]/13
=[3^6*3^(6k+3)+ 4^6*4^(6k+3)]/13
=[3^6*3^(6k+3)+ 3^6*4^(6k+3)+(4^6-3^6)*4^(6k+3)]/13
=[3^6*3^(6k+3)+ 3^6*4^(6k+3)]/13+(4^6-3^6)*4^(6k+3)/13
=3^6*[3^(6k+3)+ 4^(6k+3)]/13+(4^6-3^6)*4^(6k+3)/13
前一项整除,看后一项
4^6-3^6=(4^3-3^3)(4^3+3^3)
=(64-27)(64+27)
=(64-27)*91
=(64-27)*7*13
可见后一项也是13的倍数,所以成立