设函数f（x）=2x-cosx,【an】是公差为π/8的等差数列,f（a1）+f（a2）+..+f（a5）=5π则【f（a3）】²-a1a5=求详解

Question

设函数f（x）=2x-cosx,【an】是公差为π/8的等差数列,f（a1）+f（a2）+..+f（a5）=5π
则【f（a3）】²-a1a5=
求详解

Answer 1

∵数列{an}是公差为π/8的等差数列,
且f(a1)+f(a2)+……+f(a5)=5π
2a1-cosa1+2a2-cosa2+2a3-cosa3+2a4-cosa4+2a5-cosa5=5π
∴2(a1+a2+……+a5)-(cosa1+cosa2+……+cosa5)=5π
∴(cosa1+cosa2+……+cosa5)=0
即2(a1+a2+……+a5)=2×5a3=5π,
a3=π/2,
a1=π/4
a5=3π/4
∴[f(a3)]²-a1a5
=(2a3-cosa3)²-a1a5
=(2*π/2-cosπ/2)²-π/4*3π/4
=π²-3π²/16
=13π²/16