已知数列{An}的前n项和为Sn,点(An+2,Sn+1)在直线y=4x-5上.令Bn=An+1-2An,且A1=1(1)求数列{Bn}的通项公式(2)求数列{n·Bn}的前n项和Tn感激不尽!
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![已知数列{An}的前n项和为Sn,点(An+2,Sn+1)在直线y=4x-5上.令Bn=An+1-2An,且A1=1(1)求数列{Bn}的通项公式(2)求数列{n·Bn}的前n项和Tn感激不尽!](/uploads/image/z/4046539-67-9.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7BAn%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%E7%82%B9%28An%2B2%2CSn%2B1%29%E5%9C%A8%E7%9B%B4%E7%BA%BFy%3D4x-5%E4%B8%8A.%E4%BB%A4Bn%3DAn%2B1-2An%2C%E4%B8%94A1%3D1%281%29%E6%B1%82%E6%95%B0%E5%88%97%7BBn%7D%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F%282%29%E6%B1%82%E6%95%B0%E5%88%97%7Bn%C2%B7Bn%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8CTn%E6%84%9F%E6%BF%80%E4%B8%8D%E5%B0%BD%21)
已知数列{An}的前n项和为Sn,点(An+2,Sn+1)在直线y=4x-5上.令Bn=An+1-2An,且A1=1(1)求数列{Bn}的通项公式(2)求数列{n·Bn}的前n项和Tn感激不尽!
已知数列{An}的前n项和为Sn,点(An+2,Sn+1)在直线y=4x-5上.令Bn=An+1-2An,且A1=1
(1)求数列{Bn}的通项公式
(2)求数列{n·Bn}的前n项和Tn
感激不尽!
已知数列{An}的前n项和为Sn,点(An+2,Sn+1)在直线y=4x-5上.令Bn=An+1-2An,且A1=1(1)求数列{Bn}的通项公式(2)求数列{n·Bn}的前n项和Tn感激不尽!
(1)点(An+2,Sn+1)在直线y=4x-5上,所以:A(n+2)=4S(n+1)-5,A(n+1)=4S(n)-5,两式相减,得:A(n+2)=5A(n+1),所以A(n)=5A(n-1),所以A(n)=
A2*5^(n-2)=-5^(n-2) (n>=2)
B1=A2-2A1=-3,n>=2时,B(n)=-5^(n-1)+2*5^(n-2)=-3*5^(n-2),
(2) Tn=-3-3{2*5^0+3^5^1+4*5^2+……+n*5^(n-2)}
=-9-3{3^5^1+4*5^2+……+n*5^(n-2)}
5Tn=-15-3{2*5^1+3^5^2+4*5^3+……+n*5^(n-1)},想减,得到:
4Tn=-12+3{5^1+5^2+……+5^(n-2)}-3n*5^(n-1)=-(12n-3)/16*5^(n-1)-63/16