1+3+5+7.+(2n-1)+(2n+1)=

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1+3+5+7.+(2n-1)+(2n+1)=
1+3+5+7.+(2n-1)+(2n+1)=

1+3+5+7.+(2n-1)+(2n+1)=
设 S＝1+3+5+7.+(2n-1)+(2n+1)
倒过来写 S＝(2n+1)+（2n-1)+.+7+5+3+1
两个S相加2S＝(2n+2)+(2n+2)+(2n+2)+.(2n+2)+(2n+2)+(2n+2) 有n+1个
所以2S＝（n+1）(2n+2)＝2（n+1）^2
S＝（n+1）^2

1+3+5+7.......+(2n-1)+(2n+1)=(1+2n+1)(n+1)\2=(n+1)^2

证明不等式：(1/n)^n+(2/n)^n+(3/n)^n+.+(n/n)^n 3(n-1)(n+3)-2(n-5)(n-2) 2^n*3^n*5^(n+1)/30^n n+(n+1)+(n+2)+(n+3)+(n+4)=5n+10这道题怎么解 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, 1+(n+2)+(2n+3)+(3n+4)+(4n+5)+……((n-1)n+n)的答案 [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 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) 求证：N=(5^2)*(3^2n+1)*(2^n)-(3^n)*(6^n+2) lim(3^2n+5^n)/(1+9^n)3^2n or 5^n 求limn→无穷(1+2^n+3^n+4^n+5^n)^1/n 已知n＾2-n -1=0,则n^3-n^2-n +5= 已知(n+1)*(n+1)-n*n=2n+1 怎么解1+3+5+7.+2009+2011