数列{an}中,a1=7/2,an=3a(n-1)+(3^n)-1,(n>=2,n∈N)(1)求证{(an-1/2)/3^n}是等差数列(2)求数列{an}的通项公式an及它的前n项和sn
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![数列{an}中,a1=7/2,an=3a(n-1)+(3^n)-1,(n>=2,n∈N)(1)求证{(an-1/2)/3^n}是等差数列(2)求数列{an}的通项公式an及它的前n项和sn](/uploads/image/z/8615146-58-6.jpg?t=%E6%95%B0%E5%88%97%7Ban%7D%E4%B8%AD%2Ca1%3D7%2F2%2Can%3D3a%EF%BC%88n-1%EF%BC%89%2B%EF%BC%883%5En%EF%BC%89-1%2C%28n%3E%3D2%2Cn%E2%88%88N%29%281%29%E6%B1%82%E8%AF%81%7B%28an-1%2F2%29%2F3%5En%7D%E6%98%AF%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%EF%BC%882%EF%BC%89%E6%B1%82%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8Fan%E5%8F%8A%E5%AE%83%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8Csn)
数列{an}中,a1=7/2,an=3a(n-1)+(3^n)-1,(n>=2,n∈N)(1)求证{(an-1/2)/3^n}是等差数列(2)求数列{an}的通项公式an及它的前n项和sn
数列{an}中,a1=7/2,an=3a(n-1)+(3^n)-1,(n>=2,n∈N)(1)求证{(an-1/2)/3^n}是等差数列(2)求数列{an}的通项公式an及它的前n项和sn
数列{an}中,a1=7/2,an=3a(n-1)+(3^n)-1,(n>=2,n∈N)(1)求证{(an-1/2)/3^n}是等差数列(2)求数列{an}的通项公式an及它的前n项和sn
证明:(1)∵an=3a[n-1]+3^n-1(n≥2,n属于N*). [n-1]为下标
∴(an-1/2)/3^n-(a[n-1]-1/2)/3^[n-1]=(3a[n-1]+3^n-1-1/2)/(3^n)-(a[n-1]-1/2)/(3^[n-1])
=1
∴数列{(an-1/2)/3^n}是以(a1-1/2)/3^1=1,d=1的等差数列
(2)∵an-1/2)/3^n=1+(n-1)*1
∴an=n*3^n+1/2 n∈N+
先设bn=n*3^n,cn=1/2,另设数列{bn}的前n项和为Tn
然后用错位相减法来求Tn
Tn=1*3+2*3²+3*3³+……+n*3^n ①
3Tn= 1*3²+2*3³+……+(n-1)*3^n+n*3^(n+1) ②
②-①得Tn=(3/2n-3/4)*3^n+3/4
则Sn=Tn+1/2n=(3/2n-3/4)*3^n+3/4+1/2n n∈N+