2道高一数列题!1.已知数列{an}中,a1=2,a(n+1)=4an-3n+1,n属于N*（1）求证数列{an-n}是

1.a(n+1)=4an-3n+1=4(an-n)+(n+1)
a(n+1)-(n+1)=4[an-n]
故：{an-n}是以a1-1=1为首项,4为公比的等比数列
（2）由（1）得：an-n=4^(n-1)
an=4^(n-1)+n
Sn=(1+4+...+4^(n-1)]+(1+2+...+n)
=(4^n -1)/3 +n(n+1)/2
(3) S(n+1)-4Sn
=1-(3n^2+n-2)/2≤0
故：S(n+1)≤4Sn
2.a,b,c成等差数列,则a+c=2b
由余弦定理
cosB=(a^2+c^2-b^2)/2ac
=(3a^2+3c^2-2ac)/8ac
=1/8*(3a/c+3c/a-2)
≥1/8*(6-2)=1/2
故 0