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已知数列{An}的前n项和为Sn满足Sn=2^n%1,则数列{An...

解 ∵a(n+1)=S(n+1)-Sn Sn=2a(n+1)=2[S(n+1)-Sn] 3Sn=2S(n+1) S(n+1)/Sn=3/2 S1=1 ∴Sn=S1*(3/2)^(n-1)=(3/2)^(n-1) (n>=1)

1. 证: n=1时,S1=a1=-a1-(1/2)^0+2=-a1+1 2a1=1 a1=1/2 n≥2时, Sn=-an-(1/2)^(n-1) +2 S(n-1)=-a(n-1)-(1/2)^(n-2)+2 Sn-S(n-1)=-an-(1/2)^(n-1)+2+a(n-1)+(1/2)^(n-2)-2=-an+a(n-1)-1/2^(n-2) 2an=a(n-1)-1/2^(n-2) 等式两边同乘以2^(n-1) a...

解: (1) 设{an}公差为d,{bn}公比为q 4Sn=an·a(n+1) 4S(n+1)=a(n+1)·a(n+2) 4S(n+1)-4Sn=4a(n+1)=a(n+1)·a(n+2)-an·a(n+1) a(n+2)-an=2d=4 d=2 an=a1+(n-1)d=2+2(n-1)=2n b1b2b3=b2³=1/64 b2=1/4 b2/b1=q=(1/4)/(1/2)=1/2 bn=b1q^(n-1)=(1...

设等差数列{a[n]}的公差为d,则a[n+1]=a[1]+nd,S[n]=na[1]+(n(n-1)/2)d, 由b[n]=2(S[n+1]−S[n])S[n]−n(S[n+1]+S[n])(n∈(N^*)),得b[n]=2a[n+1]S[n]−n(2S[n]+a[n+1]) 又由b[n]=0,得 2(a[1]+nd)[na[1]+(n(...

Sn = 2(an-1) n=1, a1=2 for n≥2 an = Sn - S(n-1) =2an -2a(n-1) an = 2a(n-1) =2^(n-1) .a1 =2^n a1.b1+a2.b2+...+an.bn=(n-1).2^(n+1)+ 2 (1) n=1 a1.b1=2 b1=1 a1.b1+a2.b2+...+a(n-1).b(n-1)=(n-2).2^n + 2 (2) (1)-(2) an.bn = n2^n 2^n.b...

已知数列{a‹n›}的前n项和为S‹n›,且S‹n›=2a‹n›-1 求数列{a‹n›}的通项公式. 解:S₁=a₁=2a₁-1;∴a₁=1. S₂=a₁+a₂=2a₂-1;∴a₂=...

数列是正项数列,则an>0,Sn>0 2√Sn=an+1 4Sn=(an+1)² n=1时,4a1=4S1=(a1+1)² (a1-1)²=0 a1=1 n≥2时,4an=4Sn-4S(n-1)=(an+1)²-[a(n-1)+1]² an²-a(n-1)²-2an-2a(n-1)=0 [an+a(n-1)][an-a(n-1)]-2[an+a(n-1)...

(1)由2Sn=an2+an.①得2Sn-1=an-12+an-1.②①-②,得:2an=an2+an?an?12?an?1,∴an+an?1=an2?an?12,∴an-an-1=1,∴{an}是公差为1的等差数列,由2S1=a12+a1,得a1=1,∴an=1+(n-1)×1=n.(2)bn=2anlog 122an=-n?2n,∴Hn=-(1×2+2×22+3×23+…+n×...

∵Sn=2n-1,所以当n≥2时,an=Sn-sn-1=2n-1,又因为a1=s1=1适合上式,所以an=2n-1,故an2=4n-1,即{an2}是以1为首项,4为公比的等比数列,代入等比数列的求和公式可得其和为:13(4n-1).故选 B

解: n=1时,a1=S1=2+1+2p=2p+3 n≥2时, an=Sn-S(n-1)=2ⁿ+1+2p-(2ⁿ⁻¹+1+2p)=2ⁿ⁻¹ a(n+1)/an=2ⁿ/2ⁿ⁻¹=2 等比数列的公比q=2 a2=2²⁻¹=2 要a1是等比数列的首项, ...

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