a1=1/2,Sn=n^2an-n(n-1),求Sn和an
来源:学生作业帮助网 编辑:作业帮 时间:2024/05/15 10:34:39
a1=1/2,Sn=n^2an-n(n-1),求Sn和an
a1=1/2,Sn=n^2an-n(n-1),求Sn和an
a1=1/2,Sn=n^2an-n(n-1),求Sn和an
因an=Sn-S(n-1)
所以Sn=n²[Sn-S(n-1)]-n(n-1)
化为 [(n+1)/n]Sn-[n/(n-1)]S(n-1)=1
所以{(n+1)*Sn/n}是公差为1的等差数列
首项2S1=1
所以(n+1)*Sn/n=1+n-1=n
故Sn=n²/(n+1)
an=Sn/n²+(n-1)/n=1-1/n(n+1)
Sn=n^2An - n(n-1)
sn=n²*an-n(n-1)
an=sn-s(n-1)
sn=n²*(sn-s(n-1))-n(n-1)
a1=1/2,Sn=n^2an-n(n-1),求Sn和an
an=a1+(n-1)d 和Sn=n(a1+an)/2中an是什么意思?
an=a1+(n-1)d 和Sn=n(a1+an)/2中an是什么
a1=1,Sn=n^2an,求an
等差数列公式Sn = a1+a2+...+anSn = a1+(a1+d)+(a1+d+d)+...+[a1+(n-1)d]Sn = a1*n+[1+2+...+(n-1)]*dSn = a1*n+n*(n-1)/2*dSn = a1*n+n*(n-1)*d/2Sn = (a1+an)*n/2an = a1+(n-1)*dSn = [2a1+(n-1)*d]*n/2Sn = a1*n+n*(n-1)*d/2
数列an满足a1=1/3,Sn=n(2n-1)an,求an
Sn=na1+([n(n-1)]/2)*d=[n(a1+an)]/2【等差数列】
a1=1,n,an,Sn成等差数列,证明{Sn+n+2}是等比数列
已知数列{an}的前n项和为Sn,若a1=1/2,Sn=n^2an-n(n-1)求Sn,an
An=2An-1+2^n+2,n》2,A1=2,Sn为数列{An}的前N项和,证明Sn>n^3+n^2
等差数列求和公式Sn=n(a1+an)/2 或Sn=a1*n+n(n-1)d/2 注:an=a1+(n-1)d an=am+(n-m)*d(m小于n) 转
数列an ,a1=1,前n项和为Sn ,正整数n对应的n an Sn 成等差数列.1.证明{Sn+n+2}成等比数列,2.求{n+2/n(n+1)(1+an)}前n项和
已知a1=1,Sn=2an-2n(n>=1),则an=
Sn=n(2n-1)an,a1=1/3,求an
数列An的前n项和为Sn,已知A1=1,An+1=Sn*(n+2)/n,证明数列Sn/n是等比数列
数列{an}前n项和为sn,若sn/n=3s(n-1)/n-1(n≥2),a1=3,求an
已知数列an中,a1=-2,an+1=Sn(n∈N+),求an和Sn的表达式
Sn=2an+n a1=1 求an其中{an}的前N项和为Sn