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已知等差数列an的公差d不等于零前n项和为sn若s三等...

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(Ⅰ)证明:∵等差数列{an}的公差d不为零,Sn为其前n项和,S6=5S3,∴6a1+15d=15a1+15d,解得a1=0,∴a2=d,a3=2d,a5=4d,∵d≠0,∴a2,a3,a5成等比数列,且公比为2.(Ⅱ)解:∵a2=2,∴d=2,a2,a3,a5为等比数列{bn}的前三项,∴b1=2,b2=4,b3=8,...

(1)∵数列{an}是等差数列且s5=70,∴5a1+10d=70.①∵a2,a7,a22成等比数列,∴a72=a2?a22,即(a1+6d)2=(a1+d)(a1+21d).②由①,②解得a1=6,d=4或a1=14,d=0(舍去).∴an=4n+2.(2)证明;由(1)得Sn=2n2+4n,∴1Sn=12n2+4n=14(1n-1n+2).∴...

①③④正确,

d=2 所以Sn=n²

∵a1=1,a1、a3、a13 成等比数列,∴(1+2d)2=1+12d.得d=2或d=0(舍去),∴an =2n-1,∴Sn=n(1+2n?1)2=n2,∴2Sn+14an+3=2n2+142n+2.令t=n+1,则2Sn+14an+3=t+8t-2t=2时,t+8t-2=4,t=3时,t+8t-2=113,∴2Sn+14an+3的最小值为113.故选:D.

不知道怎么回答了 呵呵

∵a1=2,a1、a3、a13 成等比数列,∴(2+2d)2=2(2+12d).得d=4或d=0(舍去),∴an =4n-2,∴Sn=2n2,∴Sn+1612an+3=2n2+162n+2.令t=n+1,则2Sn+16an+3=t+9t-2≥6-2=4当且仅当t=3,即n=2时,∴2Sn+16an+3的最小值为4.故选:A.

(1)由题意得5a1+10d=70(a1+6d)2=(a1+d)(a1+21d)解得a1=6d=4或a1=14d=0(舍去),∴an=4n+2;(2)1Sn=12n2+4n=14(1n?1n+2),∴Tn=38?14(1n+1+1n+2),∴

∵数列{an}是以d为公差的等差数列,且a1=d,∴a2=2d,a3=3d.a12+a22+a32=14d2.又数列{bn}是公比q的等比数列,且b1=d2,∴b2=d2q,b3=d2q2.∴a12+a22+a32b1+b2+b3=14d2d2(1+q+q2)=141+q+q2∈N*.∵q是正整数,∴1+q+q2=7,解得q=2.∴S92T8=(9d+9...

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