a1=1,Sn=4a(n-1)+1 n>=2,bn=a(n+1)-2an ,Cn=1/(2^n)*an 求bn和Cn
已知数列an,bn,cn满足[a(n+1)-an][b(n+1)-bn]=cn
数列an前n项和sn,数列bn中b1=a1,bn=an-a(n-1)(n>=2),若an+sn=n(1)设cn=an-1
已知数列{an}中a1=1 a[n+1]=3an 数列{bn}的前几项和Sn=n^2+2n,设cn=an*bn,求Tn=
已知数列an=4n-2和bn=2/4^(n-1),设Cn=an/bn,求数列{Cn}的前n项和Tn
已知an=n,bn=4^n-1数列cn的通项公式cn=an*bn求cn的sn
lim(n->无穷)[(3n^2+cn+1)/(an^2+bn)-4n]=5
数列{An}的前n项和是Sn,数列{Bn},B1=A1,Bn=An-A(n-1),An+Sn=n,Cn=An-1,证{C
{an},{bn}均为d=1的等差数列a1+b1=5,设Cn=a(bn)求Cn及{Cn}的前n项和Sn
已知数列{an}中a1=1 an+1=3an 数列{bn}的前几项和Sn=n^2+2n,设cn=an*bn,求Tn=C1
已知数列{an}是an=2^n,bn=3n+1的等比数列,Cn=(3n+1)*2^n求Sn.要完整过程.
设数列{an}的前n项和为Sn=2an-4,bn=log2an,cn=1/bn^2,求证:数列{an}是等比数列?
an前n和sn且sn=2-1/2的n-1次方{bn}为等差数列a1=b1,a2*(b2-b1)=a1 求bn通项?设cn