1/[n(n+1)]+1/[(n+1)(n+2)]+1/[(n+2)(n+3)]+1/[(n+3)(n+4)]+[(n+4)(n+5)}怎么算谢谢了,

来源：学生作业学帮网编辑：学帮网时间：2024/04/29 05:41:50

1/[n(n+1)]=1/n-1/(n+1) 因此,上式=(1/n-1/(n+1))+(1/(n+1)-1/(n+2))+...+(1/(n+4)-1/(n+5)) =1/n-1/(n+5) =5/[n(n+5)]

n^(n+1/n)/(n+1/n)^n 2^n/n*(n+1) (n+1)^n-(n-1)^n=? 化简：(n+1)!/n!-n!/(n-1)! (n-1)*n!+(n-1)!*n 推导 n*n!=(n+1)!-n! 证明不等式：(1/n)^n+(2/n)^n+(3/n)^n+.+(n/n)^n 根号（n+1)+n n.(n-1). (n+2)!/(n+1)! 判断当n>1时,n*n*n>3n.( ) [3n(n+1)+n(n+1)(2n+1)]/6+n(n+2)化简 [3n(n+1)+n(n+1)(2n+1)]/6+n(n+2)化简 9题 = 101 (n+1)!- = n*n!n(n+1)!- n*n! 化简n分之n-1+n分之n-2+n分之n-3+.+n分之1 化简n分之n-1+n分之n-2+n分之n-3+.+n分之1 f(x)=e^x-x 求证(1/n)^n+(2/n)^n+...+(n/n)^n 证明[n/(n+1)]^(n+1)