设函数f(x)=3x^2+1,g(x)=2x,
现有数列{an}满足条件:对于n属于正整数,an＞0且f[(an)+1]-f(an)=g[a(n+1)+3/2]
又数列{bn}满足:bn=log(an)a(a为正数,n属于正整数)
1.求证:{an}为等比数列.
2.设p,q属于N,且p+q=5,bp=1/(1+3q),bq=1/(1+3p),求{bn}通项公式.
注:f[(an)+1]-f(an)=g[a(n+1)+3/2] an的n为下标 a(n+1)的n+1为下标
bn=log(an)a an为底数
现有数列{an}满足条件:对于n属于正整数,an＞0且f[(an)+1]-f(an)=g[a(n+1)+3/2]
又数列{bn}满足:bn=log(an)a(a为正数,n属于正整数)
1.求证:{an}为等比数列.
2.设p,q属于N,且p+q=5,bp=1/(1+3q),bq=1/(1+3p),求{bn}通项公式.
注:f[(an)+1]-f(an)=g[a(n+1)+3/2] an的n为下标 a(n+1)的n+1为下标
bn=log(an)a an为底数
