(2013•成都二模)已知数列{an}满足 an+2-an+1=an+1-an,n∈N*,且a5=π2若函数f
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(2013•成都二模)已知数列{an}满足 an+2-an+1=an+1-an,n∈N*,且a5=
π |
2 |
∵数列{an}满足an+2-an+1=an+1-an,n∈N*,
∴数列{an}是等差数列,
∵a5=
π
2,∴a1+a9=a2+a8=a3+a7=a4+a6=2a5=π
∵f(x)=sin2x+2cos2
x
2,
∴f(x)=sin2x+cosx+1,
∴f(a1)+f(a9)=sin2a1+cosa1+1+sin2a9+cosa9+1=2
同理f(a2)+f(a8)=f(a3)+f(a7)=f(a4)+f(a6)=2
∵f(a5)=1
∴数列{yn}的前9项和为9
故选C.
∴数列{an}是等差数列,
∵a5=
π
2,∴a1+a9=a2+a8=a3+a7=a4+a6=2a5=π
∵f(x)=sin2x+2cos2
x
2,
∴f(x)=sin2x+cosx+1,
∴f(a1)+f(a9)=sin2a1+cosa1+1+sin2a9+cosa9+1=2
同理f(a2)+f(a8)=f(a3)+f(a7)=f(a4)+f(a6)=2
∵f(a5)=1
∴数列{yn}的前9项和为9
故选C.
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