关于数列求和问题an=2^n+1 bn=(an -1)/(an*a(n+1)) 求bn前n项的和an= 2^n +1 就
关于数列求和问题an=2^n+1 bn=(an -1)/(an*a(n+1)) 求bn前n项的和an= 2^n +1 就
设bn=(an+1/an)^2求数列bn的前n项和Tn
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
错位相减法,数列求和an=n+1,bn=an/2^n-1,求数列bn的前n项和Tn.一轮复习,
an=3*2^(n-1),设bn=n/an求数列bn的前n项和Tn
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
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
已知bn=tan an*tan an+1,an=n+1,求数列bn前n项的和
数列{an},a1=1,an=2-2Sn,求an,若bn=n*an,求{bn}的前n项和Tn
已知数列{an}的前n项和Sn=-an-(1/2)^(n-1)+2(n为正整数).令bn=2^n*an,求证数列{bn}
a1=1.an+1=2an+2^n.bn=an/2^n-1.证明bn是等差数列、求数列的前n项和sn?
数列{an}=2n+1,使bn=1/an^2-1,求bn前n项和,