Question

求证:2^(n+2)*3^n+5n+21能被25整除

Answer 1

数学归纳

当n=1时 2^（n+2）*3^n+5n-4=25,能被25整除

当n=k时 假定2^（k+2）*3^k+5k-4能被25整除

n=k+1时 2^(k+3)*3^(k+1)+5(k+1)-4

=6*[2^（k+2）*3^k+5k-4]-25k+25,

其中2^（k+2）*3^k+5k-4能被25整除,-25k+25能被25整除,

所以2^(k+3)*3^(k+1)+5(k+1)-4能被25整除

综上所述,2^（n+2）*3^n+5n-4能被25整除