已知数列《an》满足：a1＝1.A2＝2,且an+2＝（2+cosπ）（an-1）+3,n∈nж
已知数列《an》满足：a1＝1.A2＝2,且an+2＝（2+cosπ）（an-1）+3,n∈nж
求通项公式an
求通项公式an
数学人气：461 ℃时间：2019-08-18 22:20:18
优质解答
a(1)=1,a(2)=2,a(n+2) = [2 + cos(PI)][a(n)-1] + 3 = [2-1][a(n)-1] +3= a(n) + 2,a(2n+1) = a(2n-1+2) = a(2n-1) + 2,{a(2n-1)}是首项为a(1)=1,公差为2的等差数列.a(2n-1) = 1 + 2(n-1) = 2n-1.a(2n+2) = a(2n)+2,...
我来回答
类似推荐
- 数列{an}满足a1=1,a2=2,a(n+2)=(1+cos(nπ/2)^2)an+sin(nπ/2)^2,n=1,2,3,……
- 已知数列An满足A1=1,A2=2,An+2=(1+cos*2 nπ/2)+sin2* nπ/2,则该数列的前20项的和为?
- 数列an满足a1=1,a2=2,a(n+2)=[1+cos^2(nπ/2)]an+sin^2(nπ/2)],n=1.2.3.求an的通项公式
- 数列{an}满足a1=0,a2=2,a(n+2)=(1+cos^2 *nπ/2)an+4sin^2*nπ/2,n=1,2,3.
- 数列｛an｝满足a1=1,a2=2,an+2=[1+cos（nπ/2）]an+sin（nπ/2）,n=1,2,3… （1）求a3,a4,并求数列｛an｝的通项公式 （2）设bn=a2n-1/a2n,Sn=b1+b2+…+bn 证明：当n≥6