Sn=-an-(1 2)n-1 2, bn=an2
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∵数列{an}的通项公式是an=2n−12n,∴an=1-12n,∴Sn=(1-12)+(1-14)+(1-18)+…+(1-12n)=n-(12+14+18+…+12n)=n-12[1−(12)n]
(Ⅰ):证明:∵Sn=12(n+1)(an+1)−1,∴Sn+1=12(n+2)(an+1+1)−1∴an+1=Sn+1−Sn=12[(n+2)(an+1+1)−(n+1)(an+1)]整理,得nan
平方和公式n(n+1)(2n+1)/6即1^2+2^2+3^2+…+n^2=n(n+1)(2n+1)/6(注:N^2=N的平方)证明1+4+9+…+n^2=N(N+1)(2N+1)/6证法一(归纳猜想
当公比为1时,Sn=n,数列{Sn+12}为数列{n+12}为公差为1的等差数列,不满足题意;当公比不为1时,Sn=1−qn1−q,∴Sn+12=1−qn1−q+12,Sn+1+12=1−qn+11−
an=sn-s(n-1)=13-2n(n>1)a1=s1=11所以an=13-2n(n>0)当n>1,有an-a(n-1)=-2所以an是等差数列再问:(2)求数列﹛|an|﹜前n项的和。再答:前n项
Sn=16n^2+12n-1a1=S1=27而an=Sn-S(n-1)S(n-1)=16(n-1)^2+12(n-1)-1=16n^2-20n+3所以an=32n-4n>1知a1=27不符合此式所以a
(1)∵数列{an}的前n项和为Sn,且an=12(3n+Sn)对一切正整数n成立∴Sn=2an-3n,Sn+1=2an+1-3(n+1),两式相减得:an+1=2an+3,∴an+1+3=2(an+
①当n=1时,a1=s1=32②当n≥2时,由an=sn-sn-1得an=(n2+n2)-[(n-1)2+12(n-1)]=2n-12又a1=32满足an=2n-12,所以此数列的通项公式为an=2n
证明:Sn=(π/12)*(2n^2+n)=(π/6)*(n^2)+(π/12)*n当n≥2时,S(n-1)=(π/6)*[(n-1)^2]+(π/12)*(n-1)=(π/6)*(n^2)+(π/1
Sn=12n-n^2Snmax=36Sn=12n-n^2Sn-1=12(n-1)-(n-1)^2两式相减an=12-2n+1=-2n+13数列{|An|}的前n项和Tn当n6时Tn=36+1+3+5+
Sn=12-n²an=Sn-S(n-1)=13-2n是递减数列令an6.5,即前6项为正,以后为负!故前n项和如下:(1)n≤6时Sn=12n-n²(2)n≥7时|a1+|a2|+|a
(Ⅰ)证明:把n=1代入Sn=2an+3n-12,得a1=2a1+3-12,解得a1=9,当n≥2时,an=Sn-Sn-1=(2an+3n-12)-[2an-1+3(n-1)-12]=2an-2an-
取倒数得:1/a(n+1)=(2an+1)/an=2+1/an;所以1/a(n+1)-1/an=2,又a1=1,那么1/an=2n-1,所以an=1/(2n-1)(1/an是等差数列)当n>1时bn=
(1)n≥2,sn2=(sn-sn-1)(sn-12)∴sn=sn−12sn−1+1即1sn-1sn−1=2(n≥2)∴1sn=2n-1故sn=12n−1(2)bn=sn2n+1=1(2n+1)(2n
S12>0,S1307d+24>0d>-24/7S13=(a1+a1+12d)*13/2=(2a1+12d)*13/2=13(a1+6d)=13(a1+2d+4d)=13(a3+4d)=13(12+4
等差数列求和公式:Sn=n*a1+n*(n-1)*d/2S12=12*a1+12*11*d/2=12*a1+66d>0得a1+5.5d>0S13=13*a1+13*12*s/2=13*a1+78d
(1)由sn=sn-12sn-1+1(n≥2),a1=2,两边取倒数得1Sn=1Sn-1+2,即1Sn-1Sn-1=2.∴{1sn}是首项为1S1=1a1=12,2为公差的等差数列;(2)由(1)可得
(1)Sn/n=-n+12=>Sn=-n²+12n(2)an=Sn-S(n-1)=-n²+12n+(n-1)²-12(n-1)=-2n+1+12=-2n+13所以an-a
(1)An=3(1+2^n)(2)由题知,Sn=2An+3n-12=6(2^n-1)+3nBn=(An-3)/(Sn-3n)(A(n+1)-6)=(3*2^n)/(6(2^n-1))(3(2^(n+1