an 1=((n-1)an) (n-an)
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![an 1=((n-1)an) (n-an)](/uploads/image/f/457373-29-3.jpg?t=an+1%3D%28%28n-1%29an%29+%28n-an%29)
1.an-an-1=2(n-1)-1=2(n-1)2n-2=-12n=2-12n=1n=1/22.3+(n-1)(-2)=-2n-53-2n+2=-2n-55=-5题目有错,无解.3.2+(n-1)x
(1)bn=a(2n+1)+4n-2b(n+1)=a(2n+3)+4(n+1)-2=a(2n+2+1)+4n+2=a(2n+2)-2(2n+2)+4n+2=a(2n+1+1)-2(2n+2)+4n+2
(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
(Ⅰ)∵a1=-58,an+1-an=1n(n+1),∴a2=−18,a3=124  
由于a1=-2,an+1=1−an1+an∴a2=1+a11−a1=−13,a3=1+a21−a2=12,a4=1+a31−a3=3,a5=1+a41−a4=−2=a1∴数列{an}以4为周期的数列∴
不知道你的题目是不是这样
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
(1)Sn=n^2-10nan=Sn-S(n-1)=(2n-1)-10=2n-11=>{an}是等差娄列(2)bn=(an+1)/an=(2n-10)/(2n-11)maxbn=b1=8/9minbn
A可逆,故由AA*=det(A)E知A*可逆,因此题目给出的的n-r个向量是A*的后n-r列,是线性无关的,只要证明他们是第一个方程组的解即可.由AA*=det(A)E知,A的第i(i=1,2..,r
a_(n+1)=(1+1/(n+1))^(n+1)=(1/n+1/n+...+1/n+1/(n+1))^(n+1)>[(n+1)(1/((n^n*(n+1)))开(n+1)次方根]^(n+1)(均值不
C(k,n)ak=n!/((n-k)!*k!)*(k(k+1))/2=(n-1)!/((n-k)!(k-1)!)*(n(k+1))/2=C(k-1,n-1)*n/2*(k+1)An=n/2*[C(0,
(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=
(1)证明:若an+1=an,即2an1+an=an,解得an=0或1.从而an=an-1=…a2=a1=0或1,与题设a1>0,a1≠1相矛盾,故an+1≠an成立.(2)由a1=12,得到a2=2
【方法1:强行展开a(n)表达式】1+2+……+n=n(n+1)/21^2+2^2+……+n^2=n(n+1)(2n+1)/61^3+2^3+……+n^3=n^2(n+1)^2/41^4+2^4+……
待定系数法因为a(n+1)=2an-n^2+3n设a(n+1)+p(n+1)^2+q(n+1)=2(an+pn^2+qn)展开整理得a(n+1)=2an+pn^2+(q-2p)-(p+q)与原式一一对
an=(n+1)(n+2)再问:有木有过程?再答:原式整理后得到an=(n+1)(an-1/n+1)试值:a2=(2+1)(6/2+1)=(2+1)(2x3/2+1)=12=3x4a3=(3+1)(1
2a(n+1)-an=n-2/n(n+1)(n+2)2a(n+1)-2/(n+1)(n+2)=an-1/n(n+1)[a(n+1)-1/(n+1)(n+2)]/[an-1/n(n+1)]=1/2bn=
(1)∵an+1+an=3n−54an+2+an+1=3n−51,两式相减得an+2-an=3,∴a1,a3,a5,…,与a2,a4,a6,…都是d=3的等差数列∵a1=-20∴a2=-31,①当n为
∵1=2,an+1=1+an1−an(n∈N*),∴a2=1+a11−a1=1+21−2=-3,a3=1+a21−a2=1−31+3=−12a4=1+a31−a3=1−121+12=13a5=1+a4
(1)a(n+1)/2^(n+1)=an/(an+2^n)2^(n+1)/a(n+1)=(an+2^n)/an=1+2^n/an2^(n+1)/a(n+1)-2^n/an=1所以{2^n/an}是以公