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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)由已知,得a1=S1=1×(a1-a1)2=0,∴Sn=nan2,则有Sn+1=(n+1)an+12,∴2(Sn+1-Sn)=(n+1)an+1-nan,即(n-1)an+1=nan n∈N*,∴nan+2=(n+1)an+1,两式相加得,2an+1=an+2+an n∈N*,即an+1-an+1=an+1-an n∈N*,故数列{an}是等差数列.又a1=0...

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=2an-n S(n-1)=2a(n-1)-(n-1) Sn-S(n-1)=2an-n-2a(n-1)+(n-1)=2an-2a(n-1)+1 an=2an-2a(n-1)-1 an=2a(n-1)+1 an+1=2[a(n-1)+1] [an+1]/[a(n-1)+1]=2 {an+1}为等比数列

n=1代入:a(1)=c+4 S(n)=(n+1)^2+c ……(1) n+1代入(1): S(n+1)=(n+2)^2+c ……(2) (2)-(1)化简得: a(n+1)=2n+3 即:a(n)=2n+1 由于 {an} 是等差数列,因此a(1)应该满足通项公式 所以:c+4=3 c=-1 所以 充要条件是 c=-1 必要性已证,充分性显然

an+2为等比数列Sn-1=2an-1-2(n-1) Sn-Sn-1=an=2an-2n-2an-1+2(n-1)导出an=2an-1+2 an+2=2(an-1+2)所以 an+2成等比数列

(本小题满分13分)(Ⅰ)∵4Sn=(an+1)2当n≥2时,4Sn?1=(an?1+1)2两式相减得:(an+an-1)(an-an-1-2)=0又an>0故an-an-1=2,∴{an}是以2为公差的等差数列又a1=1,∴an=2n-1.(6分)(Ⅱ)∵bn+1=abn=2bn?1,∴bn+1-1=2(bn-1)又b1-1=2≠0,∴{bn...

这个题考查数列递推式,训练了利用数学归纳法证明与自然数有关的命题,考查了学生的灵活应变能力和计算能力,是中档题. 同学这是答案哦http://gz.qiujieda.com/exercise/math/804331数列{an}的前n项和为Sn,满足Sn=2nan+1-3n^2-4n,n属于N*,且S3=15...

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