∑an=n²
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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
a(n+1)=a(n)+n+1,a(n)=a(n-1)+(n-1)+1,...a(2)=a(1)+1+1,等号两边求和.有,a(n+1)+a(n)+...+a(2)=a(n)+...+a(2)+a(1
(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
a(n+1)=an+ln[(n+1)/n]a(n+1)=an+ln(n+1)-ln(n)a(n+1)-ln(n+1)=an-ln(n)a1-ln(1)=2-0=2数列{an-ln(n)}是各项均为2的
(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  
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
an+1项应该是平方吧如果是的话,解如下:分解因式:(an+1+an)((n+1)an+1-nan)=0an+1=-an或者an+1=nan/(n+1)(1)当an+1=-an的,an=(-1)^(n
(Ⅰ)由题意可得数列{an}是首项为1,公比为3的等比数列,故可得an=1×3n-1=3n-1,由求和公式可得Sn=1×(1−3n)1−3=12(3n−1);(Ⅱ)由题意可知b1=a2=3,b3=a1
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,
a1+a2+...+an=a*n^2+bnan=4n-5/2,易知{an}为等差数列利用等差数列求和公式得:n[3/2+4n-(5/2)]/2=a*n^2+bnn(4n-1)=2a*n^2+2bn4n
(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=
应该是A(n+1)=An+2n吧~~~=>a(n+1)-an=2n所以an-a(n-1)=2(n-1)a(n-1)-a(n-2)=2(n-2)...a2-a1=2*1把左边加起来,右边加起来得到an-
马上写来再答:设级数∑An收敛于bn(An-A(n+1))=nAn-(n+1)A(n+1)-A(n+1)Sn=∑(k=1,n)[kAk-(k+1)A(k+1)-A(k+1)]=A1-(n+1)A(n+
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
(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)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