计算: n(n+1)(n+2)(n+3)+1 是哪个数的平方?

证明不等式：(1/n)^n+(2/n)^n+(3/n)^n+.+(n/n)^n 计算lim(n→∞)(1^n+2^n+3^n)^(1/n) n(n-1)(n-2)(n-3)=120,求n.求计算过程,谢谢. 计算9^n*(1/27)^n+1*3^n+2 计算：9^n*(-1/27)^n+1*3^n+2 [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 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 计算 (3A^N+2*B-2A^N*B^N-1+3B^N)*5A^N*B^N+3(N为正整数,n＞1) 化简(n+1)(n+2)(n+3) 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)...1=n 2^n/n*(n+1) lim [2^(n+2)+3^(n+3)] / [3^n-2^(n+1)] 如何计算? 计算:(3a^n+2+6a^n+1-9^n)÷3a^n-1 计算(x^(2n)+x^n+1)(x^(3n)-x^(2n)+1) 计算lim3^n-2^n+1/3^n+1+2^n=? 计算：81^n×8^2n+1÷12^3n÷3^n