(-1)^n+1·n-(-1)^n`(n-1),麻烦给化解一下,化解的答案我看不懂,答案是(-1)^n+1·（2n-1）,我想知道原因

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.( ) lim[n/(n^2+1^2)+n/(n2+2^2)+···n/(n^2+n^2)] n->无穷大 [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