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
来源:学生作业帮助网 编辑:六六作业网 时间:2024/11/23 12:02:29
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
数列{An}满足a1=1/2,a1+a2+..+an=n方an,求an
已知数列an满足an=1+2+...+n,且1/a1+1/a2+...+1/an
数列an满足a1=1/2,a1+a2+a3……an=n^2an,则an
已知数列{an}满足a1=1,a2=3,an+2=3an+1-2an求an
已知数列an'满足a1=1/2,a1+a2+a3+...+an=n^2an,求通项公式
设数列AN满足A1等于1,3(A1+a2+~+AN)=(n+2)an,求通向公式
已知数列an满足a1=0 a2=1 an=(An-1+An-2)/2 求liman
已知数列an满足a1=0 a2=1 an=(An-1+An-2)/2 求liman
数列An,满足An=An-1 + An-2,其中,A1=1,A2=1,求通项公式.
若数列{an}满足a1=1,a2=5且an+2=an+1-an,求a2000
数列{an}满足a1=3,a2=6,an+2=an+1-an,求a2013
数列{an}满足a1=3,a2=6,an+2=an+1-an,求a2008
几个数列问题.已知数列{an} a1=1,an+1=an/(1+n^2*an) 求an 已知数列{an} 满足a1=1 a1*a2*a3.*an=n^2 求an
数列an满足a1=1,a2=1/3,并且an(an-1+an+1)=2an+1an-1.则数列第2012项为?
已知数列满足a1=1/2,an+1=2an/(an+1),求a1,a2已知数列满足a1=1/2,a(n+1)=2an/(an+1),求a1,a2;证明0
已知数列{an}满足:a1+a2+a3+.+an=n^2,求数列{an}的通项an.
已知递增数列{an}满足a1=1,(2an+1)=an+(an+2),且a1,a2,a4成等比数列.求an
关于数列极限的已知数列an满足a1=0 a2=1 an=(an-1+an-2)/2 求lim(n->无穷)an