已知sn=-2 3×(1 2)^n 2 3,求an
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![已知sn=-2 3×(1 2)^n 2 3,求an](/uploads/image/f/4225228-52-8.jpg?t=%E5%B7%B2%E7%9F%A5sn%EF%BC%9D-2+3%C3%97%281+2%29%5En+2+3%2C%E6%B1%82an)
(1)数列{an}中,a1=1,前n项和Sn=n+23an,可知S2=43a2,得3(a1+a2)=4a2,解得a2=3a1=3,由S3=53a3,得3(a1+a2+a3)=5a3,解得a3=32(a
设:等差数列{an}的公差为d,通项为an=a1+(n-1)d,则:sn=a1+a2+...+an=na1+n(n-1)d/2lim(n->∞)(n*an)/Sn=lim(n->∞)[n*(a1+(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
哎呀,大家不要乱答啊,错了好多第一题Sn×SQR(S(n-1))-S(n-1)SQR(Sn)=2SQR(Sn×S(n-1)所以SQR(S(n-1)*S(n))*(SQR(S(n)-SQR(S(n-1)
∵S(n+1)-S(n)=a(n)∴a(n)=4n+2
证明:(1)由Sn=n2an−n(n−1)知,当n≥2时:Sn=n2(Sn−Sn−1)−n(n−1),…(1分)即(n2−1)Sn−n2Sn−1=n(n−1),∴n+1nSn−nn−1Sn−1=1,对
(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+
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+
(1)当n=1时,a1=S1=-14;当n≥2时,an=Sn-Sn-1=2n-8故an=−14(n=1)2n−8(n≥2)(7分)(2)由an=2n-8可知:当n≤4时,an≤0,(8分)当n≥5时,
(Ⅰ)由S1=13(a1−1),得a1=13(a1−1)∴a1=−12又S2=13(a2−1),即a1+a2=13(a2−1),得a2=14.(Ⅱ)当n>1时,an=Sn−Sn−1=13(an−1)−
(1)∵Sn=10n−n2,∴a1=S1=10-1=9.------------------(2分)当n≥2,n∈N*时,Sn−1=10(n−1)−(n−1)2=10n−n2+2n−11∴an=Sn−
因为a10=a1+9da20=a1+19d所以a20-a10=10dd=2a1=12an=12+2(n-1)=10+2nSn=242Sn=n(a1+an)/2=n[2a1+(n-1)d]/2=n(24
(I)当n=1时,a1=S1=2当n>1时,an=Sn-Sn-1=n+1,综上,数列{an}的通项公式是an=n+1(n∈N*)(II)bn=12×32−(n+1)=36×13n,b1=12,bn+1
an=n^2=n(n+1)-n=(1/3)[n(n+1)(n+2)-(n-1)n(n+1)]-(1/2)[n(n+1)-(n-1)n]Sn=a1+a2+...+an=(1/3)n(n+1)(n+2)-
(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
∵Sn=n2an,∴an+1=Sn+1-Sn=(n+1)2an+1-n2an∴an+1=nn+2an∴(1)a2=16,a3=112,a4=120(2)猜测an=1n(n+1);下面用数学归纳法证①当
(1)S1=a1=-23,∵Sn+1Sn=an-2(n≥2,n∈N),令n=2可得,S2+1S2=a2-2=S2-a1-2,∴1S2=23-2,∴S2=-34.同理可求得S3=-45,S4=-56.(
(Ⅰ)a1=3,当n≥2时,Sn−1=23an−1+1,∴n≥2时,an=Sn−Sn−1=23an−23an−1,∴n≥2时,anan−1=−2∴数列an是首项为a1=3,公比为q=-2的等比数列,∴
(Ⅰ)∵2Sn=3an-2,∴n=1时,2S1=3a1-2,解得a1=2;当n≥2时,2Sn-1=3an-1-2,∴2Sn-2Sn-1=3an-3an-1,∴2an=3an-3an-1,∴an=3an