将函数f(x)=sin ¼x•sin ¼ (x+2π)•sin ½ (x+3π)-½cos²x／2

题目

将函数f(x)=sin ¼x•sin ¼ (x+2π)•sin ½ (x+3π)-½cos²x／2
在区间（0,+∞）内的全部极值点按从小到大的顺序排成数列{an}（n∈N*）（1）求数列{an}的通项公式；（2）设bn=2nan,数列{bn}的前n项和为Tn,求Tn的表达式．

答案

(1)
f(x)=sin(x/4)•sin[(x+2π)/4]•sin [(x+3π)/2]-(1/2)cos²(x/2)
=sin(x/4)• [cos(x/4)] • [-cos(x/2)] -(1/2)cos²(x/2)
= -(1/2)sin(x/2)cos(x/2)-(1/2)cos²(x/2)
=-(1/4)sinx -(1/4)(cosx+1)
f'(x) = -(1/4)cosx + (1/4)sinx =0
tanx =1
x = kπ + π/4
an = nπ + π/4
(2)
bn = 2nan
= 2n(nπ + π/4)
Tn =b1+b2+...+bn
=2π[ (1/6)n(n+1)(2n+1) + n(n+1)/8 ]
=(π/12)n(n+1)[ 4(2n+1)+3 ]
=(π/12)n(n+1)(8n+7)