www.1862.net > 已知等比数列{An}的前n项和为Sn. 若S2=2A3+3,则...

已知等比数列{An}的前n项和为Sn. 若S2=2A3+3,则...

a1+a2=2a2+3,a1=a2+3 a1+a2+a3=2a3+3,2a2+3=a3+3,2a2=a3,q=a3/a2=2

(1)∵S2,S4,S3成等差数列,∴2S4=S2+S3,当q=1时,8a1≠2a1+3a1,舍去.当q≠1时,2?a1(1?q4)1?q=a1(1?q2)1?q+a1(1?q3)1?q,整理,得2q2-q-1=0,解得q=1(舍),或q=-12,∴数列{an}的公比q=-12.(2)∵a1-a3=3,∴a1?14a1=3,解得a1=4,∴Sn=4[1?...

(1)∵S2=a2+b2=(a+b)2-2ab=3;∵(a2+b2)(a+b)=a3+ab2+a2b+b3=a3+b3+ab(a+b),∴3×1=a3+b3-1,∴a3+b3=4,即S3=4,∵S4=(a2+b2)2-2(ab)2=7,∴S4=7;(2)∵S2=3,S3=4,S4=7,∴S2+S3=S4,∴Sn-2+Sn-1=Sn;(3)∵Sn-2+Sn-1=Sn,S2=3,S3=4...

(1)S2=a2+b2=(a+b)2-2ab=3;(2)∵(a2+b2)(a+b)=a3+ab2+a2b+b3=a3+b3+ab(a+b),∴3×1=a3+b3-1,∴a3+b3=4,即S3=4;∵S4=(a2+b2)2-2(ab)2=7,∴S4=7;(3)∵S2=3,S3=4,S4=7,∴S2+S3=S4,∴Sn-2+Sn-1=Sn;(3)∵Sn-2+Sn-1=Sn,S2=3,S...

(1)由题意可得a1q2=4a1+a1q=3解得a1=1q=2;所以an=2n?1.(2)bn=2log2a2n?log2a2n+2=2(2n?1)(2n+1)=12n?1?12n+1所以Tn=1?13+13?15+…+12n?1?12n+1=1?12n+1,因为12n+1>0,所以Tn<1.

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