为什么bn=n+1/4n^2(n+2)^2 =1/16[1/n^2 -1/(n+2)^2 ]RT,为什么(n + 1)/4n^2(n+2)^2 = 1/16 * [ 1/n^2 - 1/(n + 2)^2 ]

数列b(n+1)=bn+ 2^n.求bn. 数列求和 bn=4/(n+1)(n+2)数列求和bn=4/(n+1)(n+2),Sn=? 【大一数学分析】求极限 bn = (1-n^(-2))^ncn = n^n / (2n)!dn = n! / n^n bn=1/(2n-1)(2n+1),数列bn的前n项和为Bn,求证,Bn Bn=（2n-1)*[(4/5)的n次方] 证明Bn≤B5 数列bn=2^n/(4^n-1),证明b1+b2+b3+……+bn 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 Bn=1/(4n^2+4n),求前n项和 令bn=4/a(n+1)^2-1 (n∈N*).不等式Tn 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 已知数列{bn}满足bn=n^2/3^n,证明:bn≤4/9 bn=2/(n^2+n) 求证b1+b2+.+bn bn=（n/2+1/2）/2^n 求和 bn=1/(2n-1)(2n)求和 bn=1/(2n-1)(2n)求和