可能你的题目有误 化简通项是1/(n+1)-1/2这样 是单减列 第十个数最大
第n个数:1/(n+1)-[1+(-1)/2]*[1+(-1)^2/3]*……*[1+(-1)^(2n-1)/2n]
1个回答
相关问题
-
如何推导1/[(2n-1)(2n+1)]=1/2 *[1/(2n-1)-1/(2n+1)] 1/[(3n-1)(3n+1
-
n→∞ lim[1+1/(1+2n)+1/(2+2n)+1/(3+2n)+1/(4+2n).+1/(n+2n)]=
-
1(n-1)+2(n-2)+3(n-3)+...+(n-1)2+n1=?
-
1*N+2*(N-1)+3*(N-2)+...+N*1=1/6N(N+1)(N+2)
-
n阶行列式计算|1 2 3.n-1 n||2 1 3.n-1 n||2 3 1.n-1 n||.||2 3 4. 1 n
-
求证1-1/2+1/3-1/4……+1/(2n-1)-1/2n=1/(n+1)+1/(n+2)+……1/2n
-
1\n(n+1)+1\(n+1)(n+2)+1\(n+2)(n+3)+1\(n+3)(n+4)
-
求和:1*n+2(n-1)+3(n-2)+……+(n-2)*3+(n-1)*2+n*1 答案是n*(n+1)*(n+2)
-
1、1*n+2(n-1)+3(n-2)+……+(n-2)*3+(n-1)*2+n*1的和,如何求得1/6n(n+1)(n
-
1/2-1/n+1<1/2^2+1/3^2+……+1/n^2<n-1/n(n=2,3,4,5,6.