当n=1时,2S1=a1,解得a1=0,当n≥2时,2Sn-1=(n-1)an-1,②,由①-②可得,2an=nan-(n-1)an-1,即(n-2)an=(n-1)an-1,∴(\;a_n)/(a_(n-1))=(n-1)/(n-2),∴(a_3)/(a_2)=2/1,(a_4)/(a_3)=3/2,…,(\;a_n)/(a_(n-1))=(n-1)/(n-2),...
解:(1)在等差数列中,∵a2=11,S10=40.∴\((array)l(a_1+d=11)(10a_1+(10*9)/2d=40)(array).,即\((array)l(a_1+d=11)(a_1+9/2d=4)(array).,得a1=13,d=-2,则an=13-2(n-1)=-2n+15(n∈N•).(2)|an|=|-2n+15|=\((array)l(-2n+15,)&(1≤n≤7)(2n-15,)...
n≥2时,an=Sn-Sn-1=2an+1-(2an-1+1),∴an=2an-1,∴{an}是首项为-1,公比为2的等比数列,∴an=-2n-1.故答案为:-2n-1. 由已知条件推导出{an}是首项为-1,公比为2的等比数列,由此能求出an=-2n-1. 本题考点:数列递推式. 考点点评:本题考查数列的通项公式的求法,是中档题,解题时要认真...
(1)n=1时,a1=1.∵2Sn=3an-1,∴2Sn+1=3an+1-1,∴an+1=3an,∴an=3n-1.(2)∵bn=n⋅3n-1,∴Tn=1⋅30+2⋅31+3⋅32+…+(n-1)⋅3n-2+n⋅3n-1,3 Tn=1⋅31+2⋅32+3⋅33+…+(n-1)⋅3n-1+n⋅3n,两式相减可得-2Tn=1+31+32+…+3n-1-n⋅3n,∴Tn= 2n-1...
秒杀结论: 若an>0 ,则 S2n−1>0; 若an+am>0 ,则 Sn+m−1>0 原理: 因为S2n−1=a1+a2n−12(2n−1)=2an2(2b−1)=an(2n−1), 因此an>0 能得到 S2n−1>0 又因为 Sn+m−1=a1+an+m−12(n+m−1) ,所以由 an+am>0 ...
∴an=2n-1∴Sn=2n-1;设{bn}的公差为d,b1=-S4=-15,b9=a1=-15+8d=1,∴d=2,∴bn=2n-17;(2)cn=1(bn+16)(bn+18)=12(12n−1-12n+1),∴Wn=12[(1-13)+(13-15)+…+(12n−1-12n+1)]=12(1-12n+1)=12−14n+2 (1)由an是Sn和1的等差中项,可得Sn=2an-1,再写一...
解:(1)a2=1,由2Sn=nan,可得2a1=2S1=a1,解得a1=0,当n≥2时,由2Sn=nan,可得由2Sn-1=(n-1)an-1,上面两式相减可得2an=nan-(n-1)an-1,即为(a_n)/(a_(n-1))=(n-1)/(n-2),则an=a2•(a_3)/(a_2)•...•(a_n)/(a_(n-1))=1×2×3/2•...•(n-1)...
①-②,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 1 22an=-n?2n,∴Hn=-(1×2+2×22+3×23+…+n×2n),∴2Hn=-(22+2×23+3×24+…+n×2n+1),...
∴an=2an-2an-1,an=2an-1∴ an an−1=2,∴{an}是公比为2首项为1的等比数列,∴an=a1qn−1=2n−1.(2)∵bn=nan,=n•2n-1,∴Tn=1×1+2×21+3×22+…+n•2n−1.①,2Tn=1×21+2×22+3×23+…+n•2n.②①-②得−Tn=1+2+22+…+2n−1−n•2n= 1−2n 1...
解析 解:由an=2n-1,得Sn=a1+a2+…+an=(21-1)+(22-1)+(23-1)+…+(2n-1)=(21+22+23+…+2n)-n=2(1-2”) -n=2n+1-2-n 1-2.故选:A. 结果一 题目 若数列{an}的通项公式为an=2n-1,则数列{an}的前n项和Sn等于( ) A. 2n+1-n-2 B. 2n+1-n C. 2n+1-n+2 D. 2n+1...