数列的极限高中lim(2bn^2+4n+an^2-2n+1)/(bn+2)=1
数列的极限高中lim(2bn^2+4n+an^2-2n+1)/(bn+2)=1
讨论数列an^2+bn+2/n+1的极限
等差数列{an},{bn}的前n项和分别为An,Bn,切An/Bn=2n/3n+1,求lim(n→∞)an/bn
求数列极限lim=[(an^2+bn-1)/(4n^2-5n+1)]=1/b 求a b的值
高二的极限运算题 lim(2an+4bn)=8,lim(6an-bn)=1,求lim(3an+bn)的值 n趋向于无穷大
已知数列{an},an=2n+1,数列{bn},bn=1/2^n.求数列{an/bn}的前n项和
设bn=(an+1/an)^2求数列bn的前n项和Tn
an=3*2^(n-1),设bn=n/an求数列bn的前n项和Tn
在数列{an},{bn}中,a1=2,b1=4且an,bn,an+1成等差数列,bn,an+1,bn+1成等比数列(n∈
在数列{an},{bn}中,a1=2,b1=4,且an,bn,an+1成等差数列,bn,an+1,bn+1成等比数列(n
3.设数列{an}的前n项和Sn=2an-4(n∈N+),数列{bn}满足:bn+1=an+2bn,且b1=2.求{bn
lim(n->无穷)[(3n^2+cn+1)/(an^2+bn)-4n]=5