用数学归纳法.
证：
[2^(n+2)]×3^n +5n-4
=4×2^n×3^n+5n-4
=4×6^n+5n-4
=4×(6^n -1)+5n
n=1时,4×(6-1)+5×1=25,能被25整除.
假设当n=k(k∈N+)时,4×(6^k -1)+5n能被25整除,则当n=k+1时,
4×[6^(k+1) -1]+5(k+1)
=24×6^(k+1) -4+5k+5
=24×6^k -24 +30n -25n+25
=6×[4×(6^k-1)+5n]-25(n-1)
4×(6^k -1) +5n能被25整除,25(n-1)包含因子25,能被25整除
4×[6^(k+1) -1]+5(k+1)同样能被25整除.
k为任意正整数,因此对于任意正整数n,[2^(n+2)]×3^n +5n-4能被25整除.