数列an满足a1=1/2,a1+a2+a3……an=n^2an,则an
来源:学生作业帮助网 编辑:作业帮 时间:2024/11/19 00:22:44
xRQj@JPbbf!^@JiiGŏ .Wɪm,~vgɼ}^Dzfc"(wm e& /Kr@CUMiw
``ejJ
eɀ˭[-|u]nRL-j kt#CRM
nPg$j D7'ڜ &SU&}4vKg;2lvO߰ ]1i"泷A4CXS$ѫK;VFȌڽju(".IBnx'/n-C}7#C\-+
数列an满足a1=1/2,a1+a2+a3……an=n^2an,则an
数列an满足a1=1/2,a1+a2+a3……an=n^2an,则an
数列an满足a1=1/2,a1+a2+a3……an=n^2an,则an
s(n)=n^2a(n)
a(n+1)=s(n+1)-s(n)=(n+1)^2a(n+1)-n^2a(n)
n(n+2)a(n+1)=n^2a(n)
(n+2)a(n+1)=na(n)
(n+2)(n+1)a(n+1)=(n+1)na(n)
(n+2)(n+1)a(n+1)=(n+1)na(n)=...=(1+1)*1*a(1)=1
a(n)=1/[n(n+1)] = 1/n - 1/(n+1)
a1+a2+a3……a(n+1)=(n+1)^2an+1=n²an+a(n+1)得
(n+2)*a(n+1)=n*an即a(n+1)/an=n/(n+2)最后累乘得an=1/[(n+1)*n]
an=1/[(n+1)*n]
an=1/(n^2+n)
n^2an是n的2次方乘以数列的第n项an吗?如果是,可以用递推公式表述:
A1=1/2,An=[(n-1)^2]An-1 /[n^2-1].
数列{An}满足a1=1/2,a1+a2+..+an=n方an,求an
已知数列满足a1=1/2,an+1=2an/(an+1),求a1,a2已知数列满足a1=1/2,a(n+1)=2an/(an+1),求a1,a2;证明0
已知数列{an}中满足a1=1,a(n+1)=2an+1 (n∈N*),证明a1/a2+a2/a3+…+an/a(n+1)
(1)数列{an}中,a1=1,a2=-3,a(n+1)=an+a(n+2),则a2005=____(2)已知数列{an}满足a1=1,a1×a2×a3…an=n^2,求an.
已知数列{an}满足条件:a1=5,an=a1+a2+...a(n-1) n大于等于2,求数列{an}的通项公式
数列an满足a1=1/2,a1+a2+a3……an=n^2an,则an
已知数列an满足an=1+2+...+n,且1/a1+1/a2+...+1/an
已知正项数列{a}满足a1=1/2,且a(n+1)=an/(1+an) 1,求正项数列{a}的通项公式 2,求和:a1/1+a2/2+.2,求和:a1/1 + a2/2 +......+an/n
已知数列{an}满足a1=a,a2=b,a(n+1)=a(n+2)+an,求a2012
已知数列an'满足a1=1/2,a1+a2+a3+...+an=n^2an,求通项公式
设数列AN满足A1等于1,3(A1+a2+~+AN)=(n+2)an,求通向公式
已知数列{an}满足a(n+2)=a(n+1)-an,a1=1,a2=2,求a2005
已知数列{an}满足,a1=1,a2=2,a(n+1)=a(n+2)+an,求a2011?
已知数列{an}满足a1=1,a2=3,a(n+2)=a(n+1)-an,求S2012
已知数列{an}满足a1=4/3,2-a(n+1)=12/an+6则1/a1+2/a2+.1/an=?
已知数列{an}满足:a1=1,an=a1+2a2+3a3+``````+(n-1)a(n-1)(n大于等于2),则通项公式an是什么?
已知数列{an}满足a1=2,a(n+1)=1+an/1-an(n∈N*),a1*a2*a3*...*a2010的值为
已知数列an满足a1=1,an=a1+2a2+3a3+4a4+.(n-1)a(n-1),求通项an