已知函数f(x)=-2x+1,当x∈[An,Bn]时,f(x)的值域为[A(n+1),B(n+1)],a1=0.b1=1,数列{An},{Bn}的前n项和为Sn.Tn,设g(n)=(T1+T2+.Tn)-(S1+S2+.Sn),求g(n)
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![已知函数f(x)=-2x+1,当x∈[An,Bn]时,f(x)的值域为[A(n+1),B(n+1)],a1=0.b1=1,数列{An},{Bn}的前n项和为Sn.Tn,设g(n)=(T1+T2+.Tn)-(S1+S2+.Sn),求g(n)](/uploads/image/z/7260628-4-8.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3D-2x%2B1%2C%E5%BD%93x%E2%88%88%5BAn%2CBn%5D%E6%97%B6%2Cf%28x%29%E7%9A%84%E5%80%BC%E5%9F%9F%E4%B8%BA%5BA%28n%2B1%29%2CB%28n%2B1%29%5D%2Ca1%3D0.b1%3D1%2C%E6%95%B0%E5%88%97%7BAn%7D%2C%7BBn%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn.Tn%2C%E8%AE%BEg%28n%29%3D%28T1%2BT2%2B.Tn%29-%28S1%2BS2%2B.Sn%29%2C%E6%B1%82g%28n%29)
已知函数f(x)=-2x+1,当x∈[An,Bn]时,f(x)的值域为[A(n+1),B(n+1)],a1=0.b1=1,数列{An},{Bn}的前n项和为Sn.Tn,设g(n)=(T1+T2+.Tn)-(S1+S2+.Sn),求g(n)
已知函数f(x)=-2x+1,当x∈[An,Bn]时,f(x)的值域为[A(n+1),B(n+1)],a1=0.b1=1,数列{An},{Bn}的前n项和
为Sn.Tn,设g(n)=(T1+T2+.Tn)-(S1+S2+.Sn),求g(n)
已知函数f(x)=-2x+1,当x∈[An,Bn]时,f(x)的值域为[A(n+1),B(n+1)],a1=0.b1=1,数列{An},{Bn}的前n项和为Sn.Tn,设g(n)=(T1+T2+.Tn)-(S1+S2+.Sn),求g(n)
f(x)=-2x+1 很显然f(x)是一个单调递减函数
f(An)=B(n+1),即:-2An+1=B(n+1) (1)
f(Bn)=A(n+1),即:-2Bn+1=A(n+1) (2)
(2)-(1)得:
A(n+1)-B(n+1)=-2Bn+2An=2(An-Bn)
所以
{An-Bn}是等比数列,A1-B1=-1 公比q=2
数列{An},{Bn}的前n项和为Sn.Tn
则
Sn-Tn=(-1)(1-2^n)/(1-2)=1-2^n, Tn-Sn=2^n-1
g(n)=(T1+T2+.Tn)-(S1+S2+.Sn)
=(T1-S1)+(T2-S2)+...+(Tn-Sn)
=2-1+2^2-1+...+2^n-1
=2(1-2^n)/(1-2)-n
=2^(n+1)-n-2