求证:2^(n+2)*3^n+5n+21能被25整除

3个回答

  • 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