an=2*n+1,bn=1/(an^2-1),求数列b前n项和Tn
来源:学生作业帮助网 编辑:六六作业网 时间:2024/12/19 09:19:27
an=2*n+1,bn=1/(an^2-1),求数列b前n项和Tnan=2*n+1,bn=1/(an^2-1),求数列b前n项和Tnan=2*n+1,bn=1/(an^2-1),求数列b前n项和Tn答
an=2*n+1,bn=1/(an^2-1),求数列b前n项和Tn
an=2*n+1,bn=1/(an^2-1),求数列b前n项和Tn
an=2*n+1,bn=1/(an^2-1),求数列b前n项和Tn
答:
bn=1/(an^2-1)
=1/[(2n+1)^2-1]
=1/(4n^2+4n)
=1/[4n(n+1)]
=1/4[1/n-1/(1+n)]
所以Tn=b1+b2+...+bn
=1/4[1/1-1/2+1/2-1/3+...+1/n-1/(1+n)]
=1/4[1-1/(1+n)]
=n/(4+4n)
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
计算等差数列{an}{bn}的前n项和分别为An.Bn,且An/Bn=2n/(n+1)求limn→∞(an/bn)
已知an=n/(2^n),bn=ln(1+an)+1/2 an^2,证明,对一切n∈N*,2/(2+an)<an/bn成立
在数列an中,已知a1=2,an+1=2an/an +1,令bn=an(an -1).求证bn的前n项和
数列b=bn+an,an=1/(2^(n-1)),求bn.
a1=1,a2=2,an+2=(an+an-1)/2,n∈N+,(1)令bn=an+1-an,证明bn是等比数列
已知在直角坐标系中,An(an,0),Bn(0,bn)(n∈N*),其中数列{an},{bn}都是递增数列……已知在直角坐标系中,An(an,0),Bn(0,bn)(n∈N*),其中数列{an},{bn}都是递增数列.(1)若an=2n+1,bn=3n+1,判断直线A1B1与A2B2是否
已知:an+sn=n.1、令bn=an-1,求证:{bn}是等比数列.2、求an
设A1=2,An+1=2/An+1,Bn=|An+2/An-1|,n属于正整数,则数列{Bn}的通项公式Bn=
数学已知{an}中,Sn+an=2 1)求an 2)若{bn}中,b1=1,且b(n+1)=bn+an,求bn
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{an}是等差数列,an=2n-1,{bn}是等比数列,bn=2^(n-1)求{an/bn}前n项和Sn
an=2*3^n-1 若数列bn满足bn=an+(-1)^n*ln(an),求数列bn前n项和Sn
{an}{bn}都是等差数列,已知An/Bn(各自前n项和)=(5n+3)/(2n-1)则an/bn=?
lim(n→∞) an=2,lim(n→∞) bn=1,求lim(n→∞) (an-bn)/(an+bn)
已知等差数列{an}和{bn},他们的前n项之和为An和Bn,若An/Bn=(5n+3)/(2n-1)A9/B9
已知{an},{bn}均为等差数列,前n项的和为An,Bn,且An/Bn=2n/(3n+1),求a10/b10的值
已知an=1/n,bn^2≤bn-bn+1 (其中n属于正整数)证明(1)bn