关于数列的题目正项数列{an}满足a1=1,Sn为其前n相的和,且4Sn=(an+1)^2 (n大于等于1)求{an}的通项公式bn=an(1/2)^n,求其前n相和Tn
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![关于数列的题目正项数列{an}满足a1=1,Sn为其前n相的和,且4Sn=(an+1)^2 (n大于等于1)求{an}的通项公式bn=an(1/2)^n,求其前n相和Tn](/uploads/image/z/634430-38-0.jpg?t=%E5%85%B3%E4%BA%8E%E6%95%B0%E5%88%97%E7%9A%84%E9%A2%98%E7%9B%AE%E6%AD%A3%E9%A1%B9%E6%95%B0%E5%88%97%7Ban%7D%E6%BB%A1%E8%B6%B3a1%3D1%2CSn%E4%B8%BA%E5%85%B6%E5%89%8Dn%E7%9B%B8%E7%9A%84%E5%92%8C%2C%E4%B8%944Sn%3D%28an%2B1%29%5E2+%EF%BC%88n%E5%A4%A7%E4%BA%8E%E7%AD%89%E4%BA%8E1%EF%BC%89%E6%B1%82%7Ban%7D%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8Fbn%3Dan%EF%BC%881%2F2%EF%BC%89%5En%2C%E6%B1%82%E5%85%B6%E5%89%8Dn%E7%9B%B8%E5%92%8CTn)
关于数列的题目正项数列{an}满足a1=1,Sn为其前n相的和,且4Sn=(an+1)^2 (n大于等于1)求{an}的通项公式bn=an(1/2)^n,求其前n相和Tn
关于数列的题目
正项数列{an}满足a1=1,Sn为其前n相的和,且4Sn=(an+1)^2 (n大于等于1)
求{an}的通项公式
bn=an(1/2)^n,求其前n相和Tn
关于数列的题目正项数列{an}满足a1=1,Sn为其前n相的和,且4Sn=(an+1)^2 (n大于等于1)求{an}的通项公式bn=an(1/2)^n,求其前n相和Tn
4Sn=(an+1)^2
4S(n-1)=(a(n-1)+1)^2
[注:S和a后面的括号中为下脚标]
4an=4Sn-4S(n-1)= (an+1)^2 -(a(n-1)+1)^2
所以:(an-1)^2=(a(n-1)+1)^2
an-1=±(a(n-1)+1)
由于为{an}为正项数列,所以an-1=a(n-1)+1
an-a(n-1)=2
为等差数列,公差为2
an=1+2(n-1)=2n-1
bn=(2n-1)*(1/2)^n
Tn=1*(1/2)^1+3*(1/2)^2+5*(1/2)^3+……+(2n-3)*(1/2)^(n-1)+(2n-1)*(1/2)^n
Tn/2=1*(1/2)^2+3*(1/2)^3+5*(1/2)^4+……+(2n-3)*(1/2)^n+(2n-1)*(1/2)^(n+1)
两式相减:
Tn/2
=1/2+2*[(1/2)^2+(1/2)^3+(1/2)^4+……+(1/2)^n]-(2n-1)*(1/2)^(n+1)
=3/2-(1/2)^(n-1)-(2n-1)*(1/2)^(n+1)
=3/2-(2n+3)*(1/2)^(n+1)
Tn=3-(2n+3)*(1/2)^n