设数列{an}前n的项和为 Sn,且(3-m)Sn+2man=m+3(n∈N*).其中m为常数,m≠-3且m≠0有一步不懂请大侠解(2)若数列{an}的公比满足q=f(m)且b1=a1,bn=32 f(bn-1)(n∈N*,n≥2),求证{1 bn }为等差数列,并求bn
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![设数列{an}前n的项和为 Sn,且(3-m)Sn+2man=m+3(n∈N*).其中m为常数,m≠-3且m≠0有一步不懂请大侠解(2)若数列{an}的公比满足q=f(m)且b1=a1,bn=32 f(bn-1)(n∈N*,n≥2),求证{1 bn }为等差数列,并求bn](/uploads/image/z/10093406-14-6.jpg?t=%E8%AE%BE%E6%95%B0%E5%88%97%7Ban%7D%E5%89%8Dn%E7%9A%84%E9%A1%B9%E5%92%8C%E4%B8%BA+Sn%2C%E4%B8%94%EF%BC%883-m%EF%BC%89Sn%2B2man%3Dm%2B3%EF%BC%88n%E2%88%88N%2A%EF%BC%89%EF%BC%8E%E5%85%B6%E4%B8%ADm%E4%B8%BA%E5%B8%B8%E6%95%B0%2Cm%E2%89%A0-3%E4%B8%94m%E2%89%A00%E6%9C%89%E4%B8%80%E6%AD%A5%E4%B8%8D%E6%87%82%E8%AF%B7%E5%A4%A7%E4%BE%A0%E8%A7%A3%EF%BC%882%EF%BC%89%E8%8B%A5%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%85%AC%E6%AF%94%E6%BB%A1%E8%B6%B3q%3Df%EF%BC%88m%EF%BC%89%E4%B8%94b1%3Da1%2Cbn%3D32+f%28bn-1%29%28n%E2%88%88N%2A%2Cn%E2%89%A52%29%2C%E6%B1%82%E8%AF%81%7B1+bn+%7D%E4%B8%BA%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%2C%E5%B9%B6%E6%B1%82bn)
设数列{an}前n的项和为 Sn,且(3-m)Sn+2man=m+3(n∈N*).其中m为常数,m≠-3且m≠0有一步不懂请大侠解(2)若数列{an}的公比满足q=f(m)且b1=a1,bn=32 f(bn-1)(n∈N*,n≥2),求证{1 bn }为等差数列,并求bn
设数列{an}前n的项和为 Sn,且(3-m)Sn+2man=m+3(n∈N*).其中m为常数,m≠-3且m≠0有一步不懂请大侠解
(2)若数列{an}的公比满足q=f(m)且b1=a1,bn=32
f(bn-1)(n∈N*,n≥2),求证{1
bn
}为等差数列,并求bn.
答案是2)由b1=a1=1,q=f(m)=2m m+3 ,n∈N且n≥2时, bn=3 2 f(bn-1)=3 2 •2bn-1 bn-1+3 , (到这步就不懂了)能否解释下?⇒ 得 bnbn-1+3bn=3bn-1⇒1 bn -1 bn-1 =1 3 . ∴{1 bn }是1为首项1 3 为公差的等差数列, ∴1 bn =1+n-1 3 =n+2 3 , 故有bn=3 n+2 .
求证{1 /bn }为等差数列
并求bn
设数列{an}前n的项和为 Sn,且(3-m)Sn+2man=m+3(n∈N*).其中m为常数,m≠-3且m≠0有一步不懂请大侠解(2)若数列{an}的公比满足q=f(m)且b1=a1,bn=32 f(bn-1)(n∈N*,n≥2),求证{1 bn }为等差数列,并求bn
不好意思,你打得实在让人费解,看一下我的解题过程吧,我重做一遍:
1.(3-m)s(n)+2ma(n)=m+3
n=1,(3-m)s(1)+2ma(1)=m+3
a(1)=1
(3-m)s(n-1)+2ma(n-1)=m+3
(3+m)a(n)=2ma(n-1)
a(n)/a(n-1)=2m/(3+m)=q
a(n)=a(1)q^(n-1)=[2m/(m+3)]^(n-1)
所以数列{a(n)为等比数列
2.f(m)=2m/(m+3)
b(n)=3f(b(n-1))/2
b(n)=3*2b(n-1)/(2(b(n-1)+3)=3b(n-1)/(b(n-1)+3)
1/b(n)=1/3+1/b(n-1)
1/b(n)-1/b(n-1)=1/3=d
所以数列{1/b(n)}是等差数列