必修5的数列问题若两个等差数列{an}和{bn}的前n项和An和Bn满足关系式An/Bn=(3n+1)/(2n+3) (
必修5的数列问题若两个等差数列{an}和{bn}的前n项和An和Bn满足关系式An/Bn=(3n+1)/(2n+3) (
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
关于数列和 不等式.1.若两等差数列{an}{bn}的前n项和为 An Bn ,满足(An/Bn)=(7n+1)/4n+
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=3*2^(n-1),设bn=n/an求数列bn的前n项和Tn
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
已知两个等差数列{an},{bn}的前n项和分别是Sn,Tn,若 Sn/Tn =(2n)/(3n+1),则 an/bn=
两个等差数列{an},{bn}的前n项和分别为Sn,Tn,若Sn/Tn=2n/3n+1,求an/bn.
已知数列{an}的前n项和Sn=3×(3/2)^(n-1)-1,数列{bn}满足bn=a(n+1)/log3/2(an+
数列{an}的前n项的和Sn=2an-1(n∈N*),数列{bn}满足:b1=3,bn+1=an+bn(n∈N*).
已知等差数列an=2n-1,若数列bn=an+q^an,求数列{bn}的前n项和Sn,求详解
数列an的前n项和为Sn,Sn=4an-3,①证明an是等比数列②数列bn满足b1=2,bn+1=an+bn.求数列bn