设各项均为正数的数列{an}{bn}满足5^an,5^bn,5^an+1,成等比数列，lgbn,lg(an+1),lg(bn+1)成等差数列a1=1,b1=2,a2=3,求通项an,bn
热心网友
5^an,5^bn,5^an+1,成等比数列........an,bn,an+1成等差数列lgbn,lg(an+1),lg(bn+1)成等差数列.........bn,an+1，bn+1成等比数列an=n(n+1)/2bn=(n+1)*(n+1)/2
热心网友
(5^bn)^2=5^an*5^a(n+1)---2bn=an+a(n+1)---a(n+1)=2bn-an......(1)2lga(n+1)=lgbn+lgb(n+1)---[a(n+1)]^2=bn*b(n+1)---b(n+1)=[a(n+1)]^2/bn......(2)b1=2,a2=3 & b2=(a2)^2/b1=3^2/2=9/2a2=3,b2=9/2 & a3=2b2-a2=2*9/2-3=6b2=9,a3=6 & b3=(a3)^2/b2=6^2/9=4a3=6,b3=4 & a4=2b3-a3=2*4-6=2b3=4,a4=2 & b4=(a4)^2/b3=2^2/4=1这样得到{an}:1,3,6,2,??? {bn}:2,9/2,4,1,???数据似乎有毛病。
热心网友
(5^bn)^2=5^an*5^an+1=5^2an+12lg(an+1)=lgbn+lg(bn+1) =2lg(an+1)=lg[bn(bn+1)] =(an+1)^2=bn(bn+1)你的题目好象有点问题a1=1,b1=2,a2=3 好象不可能哦