www.1862.net > 在数列{An}中,已知A1=%1,An+1=Sn+3n%1,其中Sn...

在数列{An}中,已知A1=%1,An+1=Sn+3n%1,其中Sn...

解:(1)因为An+1=Sn+3n-1 所以An=Sn+3n-4 两式相减得An+1=2An+3...

an+1=S(n+1)-Sn=-S(n+1)*Sn 两边同除以S(n+1)*Sn得 1/Sn-1...

解:a(n+1)=Sn/3 ∴Sn=3a(n+1) 即S(n-1)=3an 两式相减得: 4...

由an+1=Sn+3n,得Sn+1?Sn=Sn+3n,∴Sn+1=2Sn+3n.则Sn+1?3n+1...

a(n+1)=KS(n+1) an=KSn 相减得a(n+1)-an=Ka(n+1) 所以a(...

(1)由an+1=Sn+(n+1)①得出n≥2时 an=Sn-1+n ②①-②得出an+1-an=a...

解:(1)Sn+1=Sn+an+1=4an1+2+an+1 ∴4an+2=4an1+2+an+...

An = Sn+1 - Sn = 2n + 1 Sn = 2(1+2+……+n) + n ...

(1)证明:由题设,anan+1=λSn-1,an+1an+2=λSn+1-1, 两式相减得an+1...

(1)a1=a,a2=-51-a,又an+1+an+2=3n-51,an+an+1=3n-54,则a...

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