已知AN=1 2N[2N 2]求[AN]的前N项和SN
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当n=1时,a1=S1=12-12=11;当n≥2时,an=Sn-Sn-1=12n-n2-[12(n-1)-(n-1)2]=13-2n.∵n=1时适合上式,∴{an}的通项公式为an=13-2n.由a
(I)由Tn=32n2-12n,易得an=3n-2代入到an+2+3log4bn=0(n∈N*)根据对数的运算性质化简bn=(14)n(n∈N*),(II)cn=an•bn=(3n−2)×(14)n,
a=1,b=1lim里面先化简,得到n+1+1/(n+1)-an,即lim[(1-a)n+1+1/(n+1)]=b所以显然有a=1,b=1没问题了吧
解题思路:已知条件是a(n+1)=3an,否则应该已知a1a2a3,前三项才能确定这个数列。解题过程:
(1)∵a1=S1=3,∴当n≥2时,an=Sn-Sn-1=2n+1,当n=1时,a1=3,∴an=2n+1…(6分)(2)当n=1时,原式=130当n≥2时,1anan+1=1(2n+1)(2n+3
a(n+1)+Sn=n^2+2nan+S(n-1)=(n-1)^2+2(n-1)两式相减得a(n+1)+Sn-an-S(n-1)=n^2+2n-(n-1)^2-2(n-1)a(n+1)+an-an=n
(1)证明:①n=1时,a1=S1=23.②n≥2时,an=Sn-Sn-1=(25n-2n2)-[25(n-1)-2(n-1)2]=27-4n,而n=1适合该式.于是{an}为等差数列.(2)因为an
①n=1时,a1=S1=-23.S2=8-50=-42,a2=S2-a1=-19,∴d=a2-a1=4,∴an=Sn-Sn-1=4n-27,an<0 得 n≤6,即数列的前6项为负
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
m²=n+2(1)n²=m+2(2)(1)-(2)m²-n²=n-m(m+n)(m-n)+(m-n)=0(m-n)(m+n+1)=0m≠nm-n≠0,要等式成立
an+2SnS(n-1)=0Sn-S(n-1)=-2SnS(n-1)1/Sn-1/S(n-1)=21/Sn-1/S1=2(n-1)1/Sn=2nSn=1/(2n)Sn.S(n+1)=1/(2n).1/
sn=a1+a2+a3+……+an=1*2^0+2*2+3*2^2+4*2^3+……+n2^(n-1)2sn=1*2+2*2^2+3*2^3+……+n*2^n两式相减得-sn=1+2+2^2+2^3+
因为Sn=-3n^2/2+205n/2,所以S[n-1]=-3(n-1)^2/2+205(n-1)/2,两式相减就得an=-3n+104.求an>0时,n
(I)a1=S1=3当n≥2时,an=Sn-Sn-1=n2+2n-[(n-1)2+2(n-1)]=2n+14,符合(II)设等比数列的公比为q,则b2=3,b4=5+7=12所以b1q=3b1q3=1
(I)当n=1时,a1=S1=4,当n≥2时,an=Sn-Sn-1=n2+2n+1-[(n-1)2+2(n-1)+1]=2n+1,又a1=4不适合上式,∴an=4,
1,这第一题是常规解法,用Sn-S(n-1)=an,你可以试一下,上下一减,得an=2n+1但是因为S(n-1)包含了第n-1项,因为n-1必须得大于等于1,所以以上必须是再n>=2时候成立,你那个好
Sn=n^2+4nS(n-1)=(n-1)^2+4(n-1)=n^2+2n-3An=S(n)-S(n-1)=2n+3
x=2n/1+n2,y=1-n2/1+n2x2+y2=[(4n^2+(1-n^2)^2]/(1+n^2)^2=(1+n^2)^2/(1+n^2)^2=1【欢迎追问,】
an=Sn-Sn-1=[1/2n2+1/2n]-[1/2(n-1)^2+1/2(n-1)]=nan=nSn=1/2n2+1/2n1/Sn=2/n(n+1)=2[1/n-1/n+1](2)记T=1/s1