1^2+3^2+5^2+...+(2n-1)^2=1/3n(4n^2-1)

来源：学生作业帮助网编辑：作业帮时间：2024/04/23 18:42:56

1^2+3^2+5^2+...+(2n-1)^2=1/3n(4n^2-1)
1^2+3^2+5^2+...+(2n-1)^2=1/3n(4n^2-1)

1^2+3^2+5^2+...+(2n-1)^2=1/3n(4n^2-1)
证明：因为1^2+2^2+...+n^2
=n(n+1)(2n+1)/6
所以1^2+2^2+...+(2n)^2
=2n(2n+1)(4n+1)/6
=n(2n+1)(4n+1)/3
2^2+4^2+...+(2n)^2
=4(1^2+2^2+...+n^2)
=4n(n+1)(2n+1)/6
=2n(n+1)(2n+1)/3
所以1^2+3^2+...(2n-1)^2
=[1^2+2^2+...+(2n)^2]-[2^2+4^2+...+(2n)^2]
=n(2n+1)(4n+1)/3-2n(n+1)(2n+1)/3
=n(2n+1)(2n-1)/3
=n(4n^2-1)/3

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 已知n＾2-n -1=0,则n^3-n^2-n +5= 求limn→无穷(1+2^n+3^n+4^n+5^n)^1/n 2^n*3^n*5^n+2/30^n