如何分解形如n^3+6n^2+11n+6 至(n+1)(n+2)(n+3)
[3n(n+1)+n(n+1)(2n+1)]/6+n(n+2)化简
证明不等式:(1/n)^n+(2/n)^n+(3/n)^n+.+(n/n)^n
1\n(n+3)+1\(n+3)(n+6)+1\(n+6)(n+9)=1\2 n+18 n为正整数,求n的值
比较2n-1和n²*n-3n²-2n+6
化简(n+1)(n+2)(n+3)
如果正整数n使得[n/2]+[n/3]+[n/4]+[n/5]+[n/6]=69,则n=
数学归纳法证明:1*n+2(n-1)+3(n-2)+…+(n-1)*2+n*1=(1/6)n(n+1)(n+2)
用数学归纳法证明:1*n+2(n-1)+3(n-2)+…+(n-1)*2+n*1=(1/6)n(n+1)(n+2)
已知888个连续正整数之和:n+(n+1)+(n+2)+(n+3)+(n+4)+(n+5)+(n+6)+(n+7)+··
计算:n(n+1)(n+2)(n+3)+1
lim[(n+3)/(n+1))]^(n-2) 【n无穷大】
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