(1)求证:4×6n+5n+1-9能被20整除;
(2)已知2n+2·3n+5n-a能被25整除,求a的最小正数值.
(1)求证:4×6n+5n+1-9能被20整除;
(2)已知2n+2·3n+5n-a能被25整除,求a的最小正数值.
(1)证明:4×6n-9=4(5+1)n-9=4(5n+
·5+1)-9 =4(5n+
·5)-5=5［4(5n-1+
)-1］是5的倍数,因此4×6n+5n+1-9是5的倍数.
又∵5n+1-9=(4+1)n+1-9=4n+1+
·4n-1+…+
·4+1-9=4·(4n+
-2)是4的倍数,因此4×6n+5n+1-9是4的倍数.
∴4×6n+5n+1-9既是4的倍数,又是5的倍数.由于4与5互质,
∴4×6n+5n+1-9能被20整除.
(2)解
:n≥2时,4×6n+5n-a=4(5+1)n+5n-a
=4(5n+C1n·5n-1+…+
=4×52(5n-2+
)+20n+4+5n-a=25×4(5n-2+
)+25n+4-a能被25整除时a=4为最小正数.
当n=1时,原式=24+5-a能被25整除时a的最小正数是4.