Sn=12 an an-1 an-2
来源:学生作业帮助网 编辑:作业帮 时间:2024/05/05 22:37:26
数列an中,a1=3,a=2an-1,∴a-1=2(an-1),∴an-1=(a1-1)*2^(n-1)=2^n,∴an=2^n+1,∴bn=2n/{(2^n+1)(2^(n+1)+1]},∴Sn=2
如果an不等于0有a(n+1)/an=2-a(n-1)a1=1,有a3=a2=1由数学归纳法可知an=1是常数列再问:不好意思是an+1+a(n+1)an-2an=0a1=1求通项再答:。。。。这个简
把1/2提取出来后,把括号里的合并1-1/3+1/3-1/5+1/5-1/7+……+1/(2n-3)-1/(2n-1)+1/(2n-1)-1/(2n+1)除了收尾两项,中间的都相互抵消所以最后结果就是
n=1,S1=a1=2,n>1,an=Sn-S(n-1)=2n,n=1时也适合,故:an=2nbn=(1/4)·1/n(n+1)4bn=1/n(n+1)=1/n-1/(n+1),所以:4Tn=[(1-
a(n+1)=2an-1a(n+1)-1=2(an-1)[a(n+1)-1]/(an-1)=2,为定值.a1-1=3-1=2数列{an}是以2为首项,2为公比的等比数列.an=2×2^(n-1)=2^
根据bn=1/(an*a(n+1)),我们知道,bn=[1/an-1/a(n+1)]/d.因此,Sn=[1/a1-1/a2+1/a2-1/a3+...+1/a(n-1)-1/an]/d=[1/a1-1
1.An+1/An=2Snn=1时,A1+1/A1=2A1->A1=1(A1>0)n>=2,An=Sn-S(n-1)∴2Sn-An=Sn+S(n-1)=1/[Sn-S(n-1)]∴S²n-S
1、n=1时,a1=S1=2a1+1a1=-1n≥2时,Sn=2an+nS(n-1)=2a(n-1)+(n-1)Sn-S(n-1)=an=2an+n-2a(n-1)-(n-1)an=2a(n-1)-1
(1)∵S4=2S2+4,∴4a1+3×42d=2(2a1+d)+4,解得d=1,(2)∵a1=−52,∴数列an的通项公式为an=a1+(n−1)=n−72,∴bn=1+1an=1+1n−72,∵函
an=2n-1Cn=1/(2n-1)(2n+1)=1/2[(1/2n-1)-(1/2n+1)]Tn=1/2(1-1/2+1/2-1/3+.1/2n-1-1/2n+1)T2011=1/2(1-1/402
a(n)=a+(n-1)d,a>0,d>0.s(n)=na+n(n-1)d/2.2s(1)=2a(1)=2a=a(1)a(n+1)=a(a+d),0=a(a+d)-2a=a(a+d-2).a+d=2.
∵数列{a[n]}满足4a[n+1]-a[n]a[n+1]+2a[n]=9∴(4-a[n])a[n+1]=9-2a[n]即:a[n+1]=(2a[n]-9)/(a[n]-4)∵a[1]=1∴a[2]=
解:an*a(n+1)+a(n+1)=2an两边同时除以an*(an+1)得:1+1/an=2/a(n+1)设:bn=1/an则:2b(n+1)=bn+12[b(n+1)-1]=bn-1[b(n+1)
a(1)=s(1)=2,a(n+1)=s(n+1)-s(n)=(n+1)^2-n^2=2n+1=2(n+1)-1.a(1)=2,n>=2时,a(n)=2n-1.b(1)=1/[a(1)a(2)]=1/
a1=S1=20-1=19,an=Sn-Sn-1=-2n+21,n≥2a1时也符合∴an=-2n+21anan+1=(-2n+21)(-2n+19)<0∴192<n<212∵n∈N∴n=10故答案为:
Sn=2n^2+nSn-1=2(n-1)^2+n-1an=Sn-Sn-1=4n-1lim[1/a1a2+1/a2a3+1/a3a4+...+1/anan+1]=lim[1/3*1/7+1/7*1/11
(Ⅰ)由bn=an-1得an=bn+1代入2an=1+anan+1得2(bn+1)=1+(bn+1)(bn+1+1)整理得bnbn+1+bn+1-bn=0从而有1bn+1−1bn=1∴b1=a1-1=
由(an-1-an)/(anan-1)=(an-an+1)/(anan+1)(n≥2),得到1/an-1/a(n-1)=1/a(n+1)-1/an{1/an}是等差数列,而且公差d=1/a2-1/a1
A1A2=4S12A2=8=>A2=4S2=6A2A3=4S2=24=>A3=6S3=12A3A4=4S3=48=>A4=8