设数列an的前n项和为sn满足2sn
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(Ⅰ)因为a1=S1,2a1=S1+2,所以a1=2,S1=2,由2an=Sn+2n知:2an+1=Sn+1+2n+1=an+1+Sn+2n+1,得an+1=sn+2n+1①,则a2=S1+22=2+
Sn=2^n-1=>an=Sn-S(n-1)=2^n-2^(n-1)=2^(n-1)bn=an+1/an=2^(n-1)+1/(2^(n-1))那么有bn-b(n-1)=(2^(n-1)-2^(n-2
∵等差数列{an}满足a3=5,a10=-9,∴a1+2d=5a1+9d=−9,解得a1=9,d=-2,∴Sn=9n+n(n−1)2×(−2)=-n2+10=-(n-5)2+25.∴n=5时,Sn取最
根据2Sn=an^2+n得到2a1=a1^2+1求得a1=1或a1=-1又因为an>0所以a1=1同理求得a2=2a3=3(2)猜想an=n证明:因为2Sn=an^2+n……①那么2Sn-1=an-1
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a1,a2+5,a3成等差数列a1+a3=2(a2+5)(1)2Sn=a(n+1)-2^(n+1)+1n=12a1=a2-4+1a2=2a1+3(2)n=22(a1+a2)=a3-8+1a3=2(a1
(1)2Sn=an^2+an2Sn-1=a(n-1)^2+a(n-1)2an=2Sn-2Sn-1=an^2-a(n-1)^2+an-a(n-1)an^2-a(n-1)^2=an+a(n-1)[an+a
由Sn=n-Sa知,an=Sn-Sn-1=1(>=2).a1=1-Sa
n=1时,S1=a1=1/2(a1+1/a1),a1=1.n=2时,S2=a1+a2=1+a2=1/2(a2+1/a2),a2=√2-1.n=3时,S3=a1+a2+a3=√2+a3=1/2(a3+1
(1)a1=1,a2=12,a3=14,a4=18(4分)猜想an=(12)n−1(6分)(2)证明:∵an=2−Sn,∴an−1=2−Sn−1(n≥2)∴an−an−1=2−Sn−(2−Sn−1),
设数列{an}的前n项和为Sn,Sn=a1(3n−1)2(对于所有n≥1),则a4=S4-S3=a1(81−1)2−a1(27−1)2=27a1,且a4=54,则a1=2故答案为2
Sn=2An-3n,Sn-Sn-1=An=2An-3n-2An-1+3(n-1),An=2An-1+3.令n=1,有A1-3=0,A1=3;B1=6(1)An=2An-1+3所以(An+3)=2(An
S(n-1)=3^(n-1)+a所以an=Sn-S(n-1)=3^n-3^(n-1)=2/3*3^n所以当n≥2的时候an一定为等比数列当n=1时s1=3+a因为a1=s1,所以a1=3+a当an为等
2Sn=an(an+1),2Sn=a(n-1)【a(n-1)+1】,an=Sn-S(n-1)得2an=an^2(平方)-a(n-1)^2+an-a(n-1).移项,平方的用平方差,因为an≠0,所以两
(1)bn=(n-1)Sn+2n-(n-2)S(n-1)-2(n-1)=(n-1)an+S(n-1)+2bn=nanan=S(n-1)+2Sn=2S(n-1)+2Sn+2=2(S(n-1)+2)得证(
/>n≥2时,an=Sn/n+2(n-1)Sn=nan-2n(n-1)S(n-1)=(n-1)an-2(n-1)(n-2)Sn-S(n-1)=an=nan-2n(n-1)-(n-1)an+2(n-1)
把a[n]+2看做一个表达式,例如为f(n),那么f(n)=a[n]+2,f(n+1)=a[n+1]+2.后面依此类推,那么表达式a[n]+2=2(a[n-1]+2),那么就相当于f(n)/f(n-1
(1)当n=1时,T1=2S1-1因为T1=S1=a1,所以a1=2a1-1,求得a1=1(2)当n≥2时,Sn=Tn-Tn-1=2Sn-n2-[2Sn-1-(n-1)2]=2Sn-2Sn-1-2n+
解题思路:考查数列的通项,考查等差数列的证明,考查数列的求和,考查存在性问题的探究,考查分离参数法的运用解题过程: