(1)
t=1
a(n+2)=a(n+1)+an
a1=a2=1
a5=5,a10=55能被5整除
若an能被5整除
则a(n+5)=a(n+4)+a(n+3)
=a(n+3)+a(n+2)+a(n+3)
=2a(n+3)+a(n+2)
=2[a(n+2)+a(n+1)]+a(n+2)
=3a(n+2)+2a(n+1)
=3[a(n+1)+an]+2a(n+1)
=5a(n+1)+3an
显然可以被5整除
(2)
t=2
a(n+2)=a(n+1)+2an
则a(n+2)+a(n+1)=2a(n+1)+2an=2[a(n+1)+an]
所以数列{a(n+1)+an}是等比数列,公比是q=2
那么a(n+1)+an=(a2+a1)*q^(n-1)=(1+1)*2^(n-1)=2^n
那么前2n项和S2n=a1+a2+a3+a4+...+a(2n-1)+a(2n)
=(a1+a2)+(a3+a4)+...+[a(2n-1)+a(2n)]
=2^1+2^3+...+2^(2n-1)
=2*(1-4^n)/(1-4)
=2(4^n-1)/3
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