Question

设函数f(x)=2x-cosx，{an}是公差为π/4的等差数列，

f(a1)+f(a2)+f(a3)=3π，则f(a1)+f(a2)+f(a3)+……+f(a10)=

Answer 1

{an}是公差为π/4的等差数列，所以：an=a1+(n-1)π/4；
f(a1)+f(a2)+f(a3)=2a1-cosa1+2（a1+π/4）-cos（a1+π/4）+2（a1+π/2））-cos（a1+π/2）
=6a1+3π/2-cos（a1+π/4）=3π——》a1=π/4，即：an=nπ/4
f(a1)+f(a2)+f(a3)+……+f(a10)=2（π/4+2π/4+...+10π/4）-（cosπ/4+cos2π/4+...+cos10π/4）
=π/2*10*（10+1）/2-cosπ/4=55π/2-v2/2。

Answer 2

因为{an}是公差为∏／4的等差数列，∴a1+a3=2a2.
f(a1)+f(a2)+f(a3)=2(a1+a2+a3)-(cosa1+cosa2+cosa3)=6a2-(cosa1+cosa2+cosa3)
∵cosa1+cosa3=cos(a2-∏／4)+cos(a2+∏／4)=2cosa2cos(-∏／2)=0(和差化积)
∴6a2-cosa2=3∏ ∵cosa2不含∏，∴cosa2=0 ∴a2=∏／2
于是a1=∏／4 an=n∏
f(a1)+f(a2)+ ….+f(a10)=2(a1+a2+…+a10)-(cosa1+cosa2+…+cosa10)
=55∏／2-√2／2