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时,丨Sn-2丨<1/n
人气:301 ℃ 时间:2019-08-21 05:35:15
解答
(1)、an+2=[1+cos(nπ/2)]an+sin(nπ/2),n∈N 因为cosnπ=2cos(nπ/2)-1=1-sin(nπ/2),所以an+2=[1+cos(nπ/2)]an+sin(nπ/2),n∈N变形为 an+2=[1+1/2+1/2cosnπ]an+1/2-1/2cosnπ=1/2[(3+cosnπ)+(1-cosnπ)],n∈N 当n为奇数时cosnπ=-1,可得an+2=1/2[(3+cosnπ)an+(1-cosnπ)]=an+2 n为奇数 此时为等差数列an=(n+1)/2 当n为偶数时cosnπ=1,可得an+2=1/2[(3+cosnπ)an+(1-cosnπ)]=2an n为偶数 此时为等比数列an=2^(n/2) (也就是根号二的n次幂) 所以通向公式为an=(n+1)/2 (n为奇数),an=2^(n/2) (n为偶数) 所以a3=(3+1)/2=2,a4=2^(4/2) =4 (2)、bn=a2n-1/a2n=[(2n-1+1)/2]/[2^(2n/2)]=n/2^n n∈N Sn=1*(1/2)+2*(1/2)^2+……+n*(1/2)^n (1/2)*Sn=1*(1/2)^2+……+(n-1)*(1/2)^n+n*(1/2)^(n+1) 相减 1/2Sn=Sn-1/2Sn=1*(1/2)+1*(1/2)^2+……+(1/2)^n-n*(1/2)^(n+1) =(1/2)*[1-(1/2)^n]/(1-1/2)-n*(1/2)^(n+1) =1-(1/2)^n-n*(1/2)^(n+1) 所以Sn=2-2*(1/2)^n-n*(1/2)^n =2-(n+2)*(1/2)^n 丨Sn-2丨=|2-(n+2)*(1/2)^n-2|=(n+2)*(1/2)^n 因为Sn是减函数,所以只要证明最大的成立则全成立 即S6成立则Sn也成立 S6=(n+2)*(1/2)^n=6/64
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