已知Sn為數(shù)列An的前n項和,且Sn=2-2An 1.求證數(shù)列 An為等比數(shù)列 2.求數(shù)列An的通項公式 3.求數(shù)列 {AnSn}的前n項和Wn
已知Sn為數(shù)列An的前n項和,且Sn=2-2An 1.求證數(shù)列 An為等比數(shù)列 2.求數(shù)列An的通項公式 3.求數(shù)列 {AnSn}的前n項和Wn
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數(shù)學(xué)人氣:545 ℃時間:2020-03-28 15:23:10
優(yōu)質(zhì)解答
an=Sn-S(n-1)=-2an+2a(n-1)3an=2a(n-1)an/a(n-1)=2/3等比數(shù)列,公比=2/3S1=2-2a1=a1,a1=2/3an=a1*q^(n-1)=(2/3)^nanSn=(2/3)^n *(2-2*(2/3)^n)=2(2/3)^n-2(2/3)^(2n)=(4/3)(2/3)^(n-1)-(8/9)(4/9)^(n-1)Wn=[(4/3)((2/...
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