数列极限 lim {[(3n^2+cn+1)/(an^2+bn)]-4n}=5,则常数a=?b=?c=?

已知lim[(3n^2+cn+1)/(an^2+bn)-4n]=5,求常数a、b、c的值 lim(n->无穷)[(3n^2+cn+1)/(an^2+bn)-4n]=5 已知数列an,bn,cn满足[a(n+1)-an][b(n+1)-bn]=cn 求数列极限lim=[(an^2+bn-1)/(4n^2-5n+1)]=1/b 求a b的值 数列极限的题目已知lim(n趋向无穷大)(5n-根号(an^2-bn+c))=2,求a,b的值 数列前N项和S=an^2+bn+c(a,b,c常数)则数列{an} 数列的极限高中lim(2bn^2+4n+an^2-2n+1)/(bn+2)=1 a,b为常数.lim（n->无穷）an^2+bn+2/2n-1=3 求a,b lim (n→∞) [(an^2+bn+c)/(2n+5)]=3,求a,b 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 已知数列an,bn,cn满足[a(n+1)-an][b(n+1)-bn]=cn 若数列an的通项公式为an=2n-1 设 设数列an前n项和为sn,an=5sn+1 bn=(4+an)/(1-an),记cn=b（2n）-b（2n-1）,求证c