Question

数学数列题，求大神，急！！！跪求！

设x∈（0,+∞），将函数f(x)＝sin(x-π/6)cos(x-π/6)+√3cos2(x-π/6)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）若an=(b1/2)+(b2/2^2)+…+(bn/2^n),求数列{bn}的前n项和Tn,

Answer 1

(1)
f(x)=sin(x-π/6)cos(x-π/6)+√3cos2(x-π/6)

= 2[(1/2)sin2(x-π/6) +(√3/2)( cos2(x-π/6) ]
= 2sin[2(x-π/6) + π/3]
= 2sin(2x)
极值

2x= (2n-1)π/2
x = (2n-1)π/4
an = (2n-1)π/4
(b1/2)+(b2/2^2)+…+(bn/2^n) = an (1)
n=1
b1/2 = π/4
b1=π/2
(b1/2)+(b2/2^2)+…+(b(n-1)/2^(n-1)) = a(n-1) (2)
(1)-(2)
bn/2^n = an - a(n-1)
=π/2
bn = π.2^(n-1)
ie
bn=π/2 ; n=1
= π.2^(n-1) ; n=2,3,4,...
b1+b2+...+bn
= π/2 + (π + 2π +...+2^(n-1)π )
= π [ 1/2 + (2^(n-1) -1 ) ]
=π [ 2^(n-1) -1/2 ]

Answer 2

题目写错了吧，前面少了个2，或者后面一项少了二分之一