an=2n-7(n∈N*),则|a1|+|a2|+-+|an|=
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1.a1=S1=2×1-a1=2-a12a1=2a1=1S2=a1+a2=a2+1=2×2-a2=4-a22a2=3a2=3/2S3=a1+a2+a3=a3+5/2=2×3-a32a3=7/2a3=7
1.an=an-1+2^nan-2*2^n=a(n-1)+2^n-2*2^nan-2*2^n=a(n-1)-2^n=a(n-1)-2*2^(n-1)可见an-2*2^n=a1-2*2^1=a1-4an
an/a(n-1)=(n+1)/(n-1)(n>=2)a(n-1)/a(n-2)=n(n-2)...a2/a1=3/1上式全部相乘an/a1=(n+1)!/2(n-1)!=n(n+1)/2,an=n(
由an=2n-7(n∈N+)得a1=2-7=-5a2=4-7=-3a3=6-7=-1a4=8-7=1由此得an是公差d=2的等差数列,且当n>=4时,an>0所以|a1|+|a2|+|a3|+……+|
(1)由已知a2=2a1+2,a3=2a2+3=4a1+7,若{an}是等差数列,则2a2=a1+a3,即4a1+4=5a1+7,得a1=-3,a2=-4,故d=-1. &nbs
An=1/(n+1)+1/(n+2)+…+1/(2n-1)+1/(2n)则An+1=1/(n+2)+1/(n+3)+…+1/(2n-1)+1/(2n)+1/(2n+1)+1/(2n+2)则An+1-A
Sn=(-1)^n*an-1/2^nS(n-1)=(-1)^(n-1)*a(n-1)-1/[2^(n-1)]两式相减得:an=(-1)^n*an-(-1)^(n-1)*a(n-1)+1/2^n.①令n
(1)当n≥2时,an=Sn-Sn-1=n(2n-1)-(n-1)(2n-3)=4n-3,当n=1时,a1=S1=1,适合.∴an=4n-3,∵an-an-1=4(n≥2),∴an为等差数列.(2)由
Sn=10n-n^2(1)S(n-1)=10(n-1)-(n-1)^2(2)(1)-(2)an=11-2nan>011-2n>02n
用a(n)=S(n)-S(n-1)的前提条件是n>=2,而n=1时利用a1=S1计算,最后再看是否符合通式.由以上思路求得a(n)=2n+5(n>=2),a(1)=9
由题意可知;an=log2n+1n+2(n∈N*),设{an}的前n项和为Sn=log223+log234+…+log2nn+1+log2n+1n+2,=[log22-log23]+[log23-lo
由an+1=an+2n可以列出以下各式a10=a9+2x9a9=a8+2x8a8=a7+2x7..a3=a2+2x2a2=a1+2x1以上各式相加可得a10=a1+1x2+2x2+.+9x2a10=9
∵an=nn2+156=1n+156n≤1439∵1n+156n≤1439当且仅当n=239时取等,又由n∈N+,故数列{an}的最大项可能为第12项或第13项又∵当n=12时,a12=12122+1
(1)证明:∵在数列{a[n]}中,已知a[n]+a[n+1]=2n(n∈N*)∴用待定系数法,有:a[n+1]+x(n+1)+y=-(a[n]+xn+y)∵-2x=2,-x-2y=0∴x=-1,y=
(2n+5)a(n+1)-(2n+7)an=4n²+24n+35=(2n+5)(2n+7)等式两边同除以(2n+5)(2n+7)a(n+1)/(2n+7)-an/(2n+5)=1a(n+1)
∵数列{an}、{bn}是等差数列,且其前n项和分别为An、Bn,由等差数列的性质得,A21=(a1+a21)×212=21a11,B21=(b1+b21)×212=21b11,∵足AnBn=7n+1
答案:a(n+1)-an=1/(2n+1)+1/(2n+2)-1/(n+1)因为an=1/(n+1)+1/(n+2)+1/(n+3)+…+1/2n所以求a(n+1)等于多少只要将(n+1)替代上式的n
(1)a1=3×1+6=9;a2=3×2+6=12a3=3×3+6=15b1=2×1+7=9b2=2×2+7=11b3=2×3+7=13∴c1=9;c2=11;c3=12;c4=13(2)解对于an=
(1)证明:∵an+1=n+2nSn,∴Sn+1−Sn=n+2nSn∴Sn+1=2n+2nSn∴Sn+1n+1=2Snn∵a1=1,∴S11=1∴数列{Snn}是以1为首项,2为公比的等比数列;(2)
(1)由于Sn=2n-an(n∈N*),所以当n=1时,S1=a1=2×1-a1,a1=1;当n=2时,S2=a1+a2=2×2-a2,a2=32当n=3时,S3=a1+a2+a3=2×3-a3,a3