设数列﹛an﹜满足a1+a2/2+a3/2^2+…+an/2^n-1=2n(1)求数列﹛an﹜的通项公式 (2)设bn=an/(an-1)(an+1-1),求数列﹛bn﹜的前n项和Sn
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![设数列﹛an﹜满足a1+a2/2+a3/2^2+…+an/2^n-1=2n(1)求数列﹛an﹜的通项公式 (2)设bn=an/(an-1)(an+1-1),求数列﹛bn﹜的前n项和Sn](/uploads/image/z/634425-33-5.jpg?t=%E8%AE%BE%E6%95%B0%E5%88%97%EF%B9%9Ban%EF%B9%9C%E6%BB%A1%E8%B6%B3a1%2Ba2%2F2%2Ba3%2F2%5E2%2B%E2%80%A6%2Ban%2F2%5En-1%3D2n%281%29%E6%B1%82%E6%95%B0%E5%88%97%EF%B9%9Ban%EF%B9%9C%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F+%EF%BC%882%EF%BC%89%E8%AE%BEbn%3Dan%2F%EF%BC%88an-1%EF%BC%89%EF%BC%88an%2B1-1%EF%BC%89%2C%E6%B1%82%E6%95%B0%E5%88%97%EF%B9%9Bbn%EF%B9%9C%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8CSn)
设数列﹛an﹜满足a1+a2/2+a3/2^2+…+an/2^n-1=2n(1)求数列﹛an﹜的通项公式 (2)设bn=an/(an-1)(an+1-1),求数列﹛bn﹜的前n项和Sn
设数列﹛an﹜满足a1+a2/2+a3/2^2+…+an/2^n-1=2n
(1)求数列﹛an﹜的通项公式 (2)设bn=an/(an-1)(an+1-1),求数列﹛bn﹜的前n项和Sn
设数列﹛an﹜满足a1+a2/2+a3/2^2+…+an/2^n-1=2n(1)求数列﹛an﹜的通项公式 (2)设bn=an/(an-1)(an+1-1),求数列﹛bn﹜的前n项和Sn
1.
n=1时,a1=2×1=2
n≥2时,
a1+a2/2+a3/2²+...+an/2^(n-1)=2n (1)
a1+a2/2+a3/2²+...+a(n-1)/2^(n-2)=2(n-1) (2)
(1)-(2)
an/2^(n-1)=2n-2(n-1)=2
an=2ⁿ
n=1时,a1=2,同样满足通项公式
数列{an}的通项公式为an=2ⁿ
2.
bn=an/[(an -1)(a(n+1) -1)]
=2ⁿ/[(2ⁿ-1)(2^(n+1)-1)]
=1/(2ⁿ-1) -1/[2^(n+1)-1]
Sn=b1+b2+...+bn
=1/(2-1) -1/(2²-1)+1/(2²-1)-1/(2³-1)+...+1/(2ⁿ-1)-1/[2^(n+1)-1]
=1 -1/[2^(n+1) -1]
a1+a2/2+a3/2^2+...+an-1/2^n-2=2(n-1),用题目已知的减去这个可得到an=2^n。bn=2^n/(2^n -1)(2^n+1 -1)。Sn=1/(2^n -1)-1/(2^n+1 -1)数学不好加上好久没算了,不知道对了没对,楼主自己仔细斟酌,仅供参考