数列的极限高中lim(2bn^2+4n+an^2-2n+1)/(bn+2)=1
来源:学生作业帮 编辑:搜搜做题作业网作业帮 分类:数学作业 时间:2024/07/16 15:32:54
数列的极限高中lim(2bn^2+4n+an^2-2n+1)/(bn+2)=1
lim(2bn^2+4n+an^2-2n+1)/(bn+2)=1
a/b=?
题目如上
我做到。a+2b=1????
lim(2bn^2+4n+an^2-2n+1)/(bn+2)=1
a/b=?
题目如上
我做到。a+2b=1????
![数列的极限高中lim(2bn^2+4n+an^2-2n+1)/(bn+2)=1](/uploads/image/z/4826915-35-5.jpg?t=%E6%95%B0%E5%88%97%E7%9A%84%E6%9E%81%E9%99%90%E9%AB%98%E4%B8%ADlim%282bn%5E2%2B4n%2Ban%5E2-2n%2B1%29%2F%28bn%2B2%29%3D1)
lim(2bn^2+4n+an^2-2n+1)/(bn+2)=1,
括号里分子分母同时除以n:lim(2bn+4+an-2+(1/n)/ (b+2/n))=1
当n趋于无穷时,1/n=2/n=0;要是方程成立,则:2b+a=0,(4-2)/b=1
因此得:b=2,a=-4.
括号里分子分母同时除以n:lim(2bn+4+an-2+(1/n)/ (b+2/n))=1
当n趋于无穷时,1/n=2/n=0;要是方程成立,则:2b+a=0,(4-2)/b=1
因此得:b=2,a=-4.
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