一、f(x)=2√3sinxcosx+2cos²x-1=√3sin2x+cos2x=2sin(2x+π/6)
(1)最小正周期為π,單調(diào)遞增區(qū)間-π/2+2kπ≤2x+π/6≤π/2+2kπ,即-π/3+kπ≤x≤π/6+kπ
(2)分為兩個區(qū)間[0,π/6]為單調(diào)遞增,(π/6,π/2]為單調(diào)遞減,[π/6,2π/3]為單調(diào)遞減區(qū)間
f(0)=1,f(π/6)=2,f(π/2)=-1,所以最大值為2,最小值為-1
二、(1)(bn+1)=(an+1)+3,(bn+1)/bn=2an+3+3/an+3=2,所以{bn}為q=2的等比數(shù)列
(2)(b1)=(a1)+3=2,bn=2*2^(n-1)=2^n,an=bn-3=2^n-3
(3)cn=n*2^n,sn=1*2¹+2*2²+3*2³+…+n*2^n ①
2sn=1*2²+2*2³+…+n*2^(n+1) ②
②-① sn=n*2^(n+1)-(1*2¹+1*2²+1*2³+…+1*2^n)=n*2^(n+1)-[2^(n+1)-2]=(n-1)2^(n+1)+2