设数列{an}的首相a1=1,前n项和Sn满足关系式:3tSn-(2t+3)S(n-1)=3t(t>0,n=2,3,4,…)(1)求证数列{an}是等比数列(要有推理过程);(2)设数列{an}的公比为f(t),做数列{bn},使b1=1,bn=f(1/b(n-1
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![设数列{an}的首相a1=1,前n项和Sn满足关系式:3tSn-(2t+3)S(n-1)=3t(t>0,n=2,3,4,…)(1)求证数列{an}是等比数列(要有推理过程);(2)设数列{an}的公比为f(t),做数列{bn},使b1=1,bn=f(1/b(n-1](/uploads/image/z/5943406-22-6.jpg?t=%E8%AE%BE%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E9%A6%96%E7%9B%B8a1%3D1%2C%E5%89%8Dn%E9%A1%B9%E5%92%8CSn%E6%BB%A1%E8%B6%B3%E5%85%B3%E7%B3%BB%E5%BC%8F%EF%BC%9A3tSn%EF%BC%8D%EF%BC%882t%2B3%EF%BC%89S%EF%BC%88n%EF%BC%8D1%EF%BC%89%3D3t%28t%EF%BC%9E0%2Cn%3D2%2C3%2C4%2C%E2%80%A6%29%EF%BC%881%EF%BC%89%E6%B1%82%E8%AF%81%E6%95%B0%E5%88%97%7Ban%7D%E6%98%AF%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%EF%BC%88%E8%A6%81%E6%9C%89%E6%8E%A8%E7%90%86%E8%BF%87%E7%A8%8B%EF%BC%89%EF%BC%9B%EF%BC%882%EF%BC%89%E8%AE%BE%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%85%AC%E6%AF%94%E4%B8%BAf%28t%29%2C%E5%81%9A%E6%95%B0%E5%88%97%7Bbn%7D%2C%E4%BD%BFb1%3D1%2Cbn%3Df%281%2Fb%EF%BC%88n%EF%BC%8D1)
设数列{an}的首相a1=1,前n项和Sn满足关系式:3tSn-(2t+3)S(n-1)=3t(t>0,n=2,3,4,…)(1)求证数列{an}是等比数列(要有推理过程);(2)设数列{an}的公比为f(t),做数列{bn},使b1=1,bn=f(1/b(n-1
设数列{an}的首相a1=1,前n项和Sn满足关系式:3tSn-(2t+3)S(n-1)=3t(t>0,n=2,3,4,…)(1)求证
数列{an}是等比数列(要有推理过程);(2)设数列{an}的公比为f(t),做数列{bn},使b1=1,bn=f(1/b(n-1))(n=2,3,4,…),求数列{bn};(3)求和:b1b2-b2b3+b3b4-b4b5+b(2n-1)b2n-b(2n+1)
设数列{an}的首相a1=1,前n项和Sn满足关系式:3tSn-(2t+3)S(n-1)=3t(t>0,n=2,3,4,…)(1)求证数列{an}是等比数列(要有推理过程);(2)设数列{an}的公比为f(t),做数列{bn},使b1=1,bn=f(1/b(n-1
1.
Sn=(2t+3)/3t*S(n-1)+1
S(n-1)=(2t+3)/3t*S(n-2)+1
两式相减得an=(2t+3)/3t*a(n-1)
所以{an}为等比数列,公比q=(2t+3)/3t
2.
f(t)=(2t+3)/3t
bn=[2/b(n-1)+3]/[3/b(n-1)]
化简得bn-b(n-1)=2/3
所以bn=1+2(n-1)/3=(2n+1)/3
3.算式应该抄错了吧,最后一项是不是 -b(2n)*b(2n+1)
b(2n-1)b2n=(4n+1)(4n-1)/9=(16n^2-1)/9
原式=∑(-1)^(k-1)*(16k^2-1)/9 k=1,2,3,……,2n-1,2n
所以 原式=(16/9)∑(-1)^(k-1)*k^2 k=1,2,3,……,2n-1,2n
∑(-1)^(k-1)*k^2 =1-2^2+3^2-4^2+……+(2n-1)^2-(2n)^2
=1+2^2 +3^2+……+(2n)^2-2*[2^2+4^2+6^2+……+(2n)^2]
=2n(2n+1)(4n+1)/6-2^3*[1+2^2 +3^2+……+(n)^2]
=n(2n+1)(4n+1)/3-8n(n+1)(2n+1)/6
=-n(n+1)
(3)设cn=b(2n-1)b2n-b2nb(2n+1)
cn=-16/9n-4/9
cn的前n项和为b1b2-b2b3+b3b4-b4b5+...+b(2n-1)b2n-b2nb(2n+1)=
-8/9n^2-4/9n