设数列{an},{bn}满足a1=1/2,2na(n+1)=(n+1)an,且{bn}=ln(1+an)+1/2an^2
1个回答
a(n+1)/(n+1)=1/2*a(n)/n
所以a(n)/n是等比数列,又a(1)/1=1/2
所以a(n)/n=1/2^n,a(n)=n/2^n
相关问题
设数列{An}和{bn}满足A1=1/2,2nA(n+1)=(n+1)An,且Bn=ln(1+An)+1/2(An)2,
已知数列{an}满足:an+an+1=2an+2,且a1=1,a2=2,n∈N* 一:设bn=an+1-an ,证明bn
(2010•莒县模拟)设数列{an}为等比数列,数列{bn}满足bn=na1+(n-1)a2+…+2an-1+an,n∈
数列{an}满足a1=2,an+1=2^(n+1)*an/((n+1/2)*an+2^n),(1)设bn=2^n/an,
设数列{an}为等比数列,数列{bn}满足bn=na1+(n-1)a2+…+2an-1+an,n∈N*,已知b1=m,b
数列{an}{bn}中,a 1=1,b1=2,且an+1+(−1)nan=bn,n∈N*,设数列{an}{bn
数列{an}满足a1=1,a2=2,an=1/2(an-1+an-2)(n=3,4...),数列{bn}满足bn=an+
数列{an}与{bn}满足关系:a1=2,a(n+1)=(an^2+1)/2an,bn=(an+1)/(an-1).(n
数列{an}满足a(n+1)=3an-2/2an-1,且a1=2.(1)设bn=1/an-1,求证{bn}为等差数列.(
在数列{an}中,a1=1,an+1=(1+1/n)an+(n+1)∕2n 设bn=an/n,求证bn+1-bn=1/2