bn=1/(2^n +1)(2^(n+1)+1),Tn=b1+b2+……+2^(n-1)bn,求证Tn小于1/6是Tn=b1+2b2+……+2^(n-1)bn

数列b(n+1)=bn+ 2^n.求bn. bn=1/(2n-1)(2n+1),数列bn的前n项和为Bn,求证,Bn 若数列bn满足bn=n^2/2^(n+1),证明bn 设数列{bn}满足bn=n^2/2^(n+1),证明:bn bn+1=bn+2n-1 bn=-1 求bn通项 bn=2^(2n-1)-2n,求{bn}的前n项和Tn bn=(n+1)2n,求数列{bn/1}的前n项和Tn 数列{bn}满足bn=(2n-1)/3^n,求前n项和,Tn 已知an=1/n,bn^2≤bn-bn+1 (其中n属于正整数)证明(1)bn 已知an=1/n,bn^2≤bn-bn+1 (其中n属于正整数)证明(1)bn 数列bn的通项公式为bn=2/n*(n-1),求bn的前n项和. bn=1/【2n】怎么求和 【大一数学分析】求极限 bn = (1-n^(-2))^ncn = n^n / (2n)!dn = n! / n^n 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 已知数列{ bn } 满足2b(n+1)= bn + 1/bn ,且bn>1,求{bn}通项公式 bn=1/(2n-1)(2n)求和 bn=1/(2n-1)(2n)求和 bn=（n/2+1/2）/2^n 求和