数列an,a1=a2=1,a（n+2）=a（n+1）+tan.（1）t=1时,写出2个能被5整除的项；证明an能被5整除

(1)t=1a(n+2)=a(n+1)+ana1=a2=1a5=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=2a(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如果不懂,请Hi我,祝学习愉快!