已知数列{an}{bn}满足,对任意n属于正整数,an,bn,an+1成等差数列,且an+1=根号下bnbn+1 1.证
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已知数列{an}{bn}满足,对任意n属于正整数,an,bn,an+1成等差数列,且an+1=根号下bnbn+1 1.证明:根号bn是等差
已知数列{an}{bn}满足,对任意n属于正整数,an,bn,an+1成等差数列,且an+1=根号下bnbn+1
1.证明:根号bn是等差数列
2.设a1=1,a2=2,求{an}{bn}的通项公式
主要是第二问
已知数列{an}{bn}满足,对任意n属于正整数,an,bn,an+1成等差数列,且an+1=根号下bnbn+1
1.证明:根号bn是等差数列
2.设a1=1,a2=2,求{an}{bn}的通项公式
主要是第二问
an,bn,a(n+1)成等差数列
2bn=an+a(n+1)
a(n+1)=√[bnb(n+1)]
an=√[bnb(n-1)]
2bn=√[bnb(n+1)]+√[bnb(n-1)]
2(√bn)^2=√[bnb(n+1)]+√[bnb(n-1)](已知数列{an}{bn}满足,对任意n属于正整数)
2√bn=√(n+1)+√b(n-1)
所以√bn是等差数列
再问: 第二问。。快啊。就要下了
再答: 2b1=a1+a2 2b1=1+2=3 b1=3/2 a2=√[b2*b1] 2=√[b2*3/2] b2=8/3 d=√b2-√b1 =√(8/3)-√(3/2) =2√6/3-√6/2 =√6/6 √bn=√b1+(n-1)d =√(3/2)+(n-1)√6/6 =n+√6/2-√6/6 =n+√6/3 bn=(n+√6/3)^2 b(n-1)=(n-1+√6/3)^2 an=√[bnb(n-1)] =√[(n+√6/3)^2(n-1+√6/3)^2] =(n+√6/3)(n-1+√6/3)
2bn=an+a(n+1)
a(n+1)=√[bnb(n+1)]
an=√[bnb(n-1)]
2bn=√[bnb(n+1)]+√[bnb(n-1)]
2(√bn)^2=√[bnb(n+1)]+√[bnb(n-1)](已知数列{an}{bn}满足,对任意n属于正整数)
2√bn=√(n+1)+√b(n-1)
所以√bn是等差数列
再问: 第二问。。快啊。就要下了
再答: 2b1=a1+a2 2b1=1+2=3 b1=3/2 a2=√[b2*b1] 2=√[b2*3/2] b2=8/3 d=√b2-√b1 =√(8/3)-√(3/2) =2√6/3-√6/2 =√6/6 √bn=√b1+(n-1)d =√(3/2)+(n-1)√6/6 =n+√6/2-√6/6 =n+√6/3 bn=(n+√6/3)^2 b(n-1)=(n-1+√6/3)^2 an=√[bnb(n-1)] =√[(n+√6/3)^2(n-1+√6/3)^2] =(n+√6/3)(n-1+√6/3)
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