a1=a2=1,a(n+2)=a(n+1)+tan若t=1时,证明an能被5整除,a(n+5)也能被5整除若t=2时,求Sn如没有时间,希望能给出思路

Question

a1=a2=1,a(n+2)=a(n+1)+tan
若t=1时,证明an能被5整除,a(n+5)也能被5整除
若t=2时,求Sn
如没有时间,希望能给出思路

Answer 1

1,a3=2,a4=3,a5=5
a(n+5)=a(n+4)+a(n+3)=a(n+3)+a(n+2)+a(n+2)+a(n+1)
=a(n+2)+a(n+1）+2a(n+1)+2an+a(n+1)
=a(n+1)+an+4a(n+1)+2an
=5a(n+1)+3an
an能被5整除,a(n+5)也能被5整除
（2）t=2,a(n+2)+a(n+1)=2[a(n+1)+an]
{a(n+1)+an}是首项a2+a1=2 公比2的等比数列.
a(n+1)+an=2^n
an+a(n-1)=2^(n-1)
.
a4+a3=2^3
a3+a2=2^2
a2+a1=2^1
相加sn-a1+s(n-1)=2^1+2^2+...+2^(n-1)
sn+s(n-1)=a1+2^1+2^2+...+2^(n-1)=2^n-1
2sn-an=2^n-1 a(n) = 1/3*[2^n - (-1)^n]