若t=1数列{an}中,若a1=a2=1,a(n+2)=a(n+1)=tan证明:若an能被5整除,则a(n+5)也能被5整除

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若t=1数列{an}中,若a1=a2=1,a(n+2)=a(n+1)=tan证明:若an能被5整除,则a(n+5)也能被5整除
若t=2,求{an}的前n项和Sn

1,a(n+5)=a(n+4)+a(n+3)=2a(n+3)+a(n+2)=3a(n+2)+2a(n+1)=5a(n+1)+3an;由于5a(n+1)能被5整除,如果an能被5整除,那么a(n+5)也能被5整除.2,a(n+2)=a(n+1)+2an,两边各加a(n+1),得a(n+2)+a(n+1)=2...