数列an满足a1=1,a2=2,an+2=(cos^2×nπ/2)an+sin^2×nπ/2,则2013=() A.2013 B.3019 C.2 D.1
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数列an满足a1=1,a2=2,an+2=(cos^2×nπ/2)an+sin^2×nπ/2,则2013=()A.2013B.3019C.2D.1数列an满足a1=1,a2=2,an+2=(cos^2
数列an满足a1=1,a2=2,an+2=(cos^2×nπ/2)an+sin^2×nπ/2,则2013=() A.2013 B.3019 C.2 D.1
数列an满足a1=1,a2=2,an+2=(cos^2×nπ/2)an+sin^2×nπ/2,则2013=() A.2013 B.3019 C.2 D.1
数列an满足a1=1,a2=2,an+2=(cos^2×nπ/2)an+sin^2×nπ/2,则2013=() A.2013 B.3019 C.2 D.1
是否求S2013
∵a1=1,a2=2
a(n+2)=cos²(nπ/2)an+sin²(nπ/2)
∴a3=cos²(π/2)a1+sin²(π/2)=1
a5=cos²(5π/2)a3+sin²(5π/2)=1
.
a(2k-1)=1
a4=cos²(π)a2+sin²(π)=2
a6=cos²(2π)a4+sin²(2π)=2
.
a2k=2
∴S2013
=(a1+a3+.+a2013)+(a2+a4+.+a2012)
=1007+1006*2=3019
选C
选D
因为当n取偶数时cos^2×nπ/2=1 sin^2×nπ/2=0
当n取奇数时cos^2×nπ/2=0 sin^2×nπ/2=1
所以原数列为1,2,1,2,1,2,1,2,....
奇数项为1,偶数项为2
所以第2013项为1
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