大师作文网已知数列{an},其首项为a1(a1≠0且为常数),前n项和Sn满足：对任意的r,t∈N,都有Sr：St=r^2:t^2
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已知数列{an},其首项为a1(a1≠0且为常数),前n项和Sn满足：对任意的r,t∈N,都有Sr：St=r^2:t^2(1)判断{an}是否为等差数列?并证明你的结论（2）若数列{abn}是等比数列,且b1=2,b2=5,求数列{bn}的通项公式
(1)证明:∵Sr/St =（r/t）²
对于r=n,t=1时同样成立
S(n)/S(1) = n^2,
S(n) = n^2S(1)=n^2a(1),
S(n+1) = (n+1)^2a(1),
a(n+1) = S(n+1)-S(n) = a(1)[(n+1)^2 - n^2] = a(1)[2n+1],
a(n) = a(1)[2n-1],
a(n+1) - a(n) = a(1)[2n+1 - 2n+1] = 2a(1),
{a(n)}是首项为a(1),公差为2a(1)的等差数列.
(2)a(n) = a(1) + 2a(1)(n-1) = a(1)[2n-1],n=1,2,...
∴a1=1,an=2n-1
bn=a(b(n-1))=2b(n-1)-1
bn-1=2(b(n-1)-1),即：bn-1是公比为2的等比数列
bn-1=(b1-1)*2^(n-1)=2^n
bn=2^n+1
tn=b1+b2+.+bn=2+1+4+1+.+2^n+1
=2+4+.+2^n+n=2^(n+1)+n-2
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对于r=n,t=1时同样成立
S(n)/S(1) = n^2,
S(n) = n^2S(1)=n^2a(1),
S(n+1) = (n+1)^2a(1),
a(n+1) = S(n+1)-S(n) = a(1)[(n+1)^2 - n^2] = a(1)[2n+1],
a(n) = a(1)[2n-1],
a(n+1) - a(n) = a(1)[2n+1 - 2n+1] = 2a(1),
{a(n)}是首项为a(1),公差为2a(1)的等差数列.
(2)a(n) = a(1) + 2a(1)(n-1) = a(1)[2n-1],n=1,2,...
∴a1=1,an=2n-1
bn=a(b(n-1))=2b(n-1)-1
bn-1=2(b(n-1)-1),即：bn-1是公比为2的等比数列
bn-1=(b1-1)*2^(n-1)=2^n
bn=2^n+1
tn=b1+b2+.+bn=2+1+4+1+.+2^n+1
=2+4+.+2^n+n=2^(n+1)+n-2
明教为您解答,
如若满意,请点击[满意答案];如若您有不满意之处,请指出,我一定改正!
希望还您一个正确答复!
祝您学业进步!
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