下列计算能否化简?设f(n)=[(n+1)n+n(n-1)+(n-1)(n-2)+...+3*2+2*1]/2 (n为正

证明不等式：(1/n)^n+(2/n)^n+(3/n)^n+.+(n/n)^n 当n为正偶数,求证n/(n-1)+n(n-2)/(n-1)(n-3)+...+n(n-2).2/(n-1)(n-3).. 2^n/n*(n+1) 计算：n(n+1)(n+2)(n+3)+1 设f(x)=2^x/(2^x+根号2),求f(1/n)+f(2/n)+f(3/n)+.+f(n/n)(n为自然数) [3n(n+1)+n(n+1)(2n+1)]/6+n(n+2)化简 (1/2)谢谢解答下列问题祝中秋快乐.设f(n)=(1/n+1)+(1/n+2)+(1/n+3)+…+(1/2n)(n 1 + (n + 1) + n*(n + 1) + n*n + (n + 1) + 1 = 2n^2 + 3n + 3 设n∈N,n>1.求证：logn (n+1)>log(n+1) (n+2) 化简(n+1)(n+2)(n+3) 证明(1+2/n)^n>5-2/n（n属于N+,n>=3) 已知递推公式f(n)=(n-1)(n-2)[f(n-2)+f(n-3)+(n-3)*f(n-4)] (n>4)求通项公式