1^3+2^3^+3^3+.+n^3

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1^3+2^3^+3^3+.+n^31^3+2^3^+3^3+.+n^31^3+2^3^+3^3+.+n^3证明1^3+2^3+3^3+...+n^3=(1+2+3+...+n)^2=[n(n+1)/

1^3+2^3^+3^3+.+n^3
1^3+2^3^+3^3+.+n^3

1^3+2^3^+3^3+.+n^3
证明1^3+2^3+3^3+...+n^3=(1+2+3+...+n)^2=[n(n+1)/2]^2 n^4-(n-1)^4 =[n^2-(n-1)^2][n^2+(n-1)^2] =(2n-1)(2n^2-2n+1) =4n^3-6n^2+4n-1 2^4-1^4=4*2^3-6*2^2+4*2-1 3^4-2^4=4*3^3-6*3^2+4*3-1 4^4-3^4=4*4^3-6*4^2+4*4-1 .n^4-(n-1)^4=4n^3-6n^2+4n-1 各等式全部相加 n^4-1^4=4*(2^3+3^3+...+n^3)-6*(2^2+3^2+...+n^2)+4(2+3+4+...+n)-(n-1) n^4-1^4=4*(1^3+2^3+3^3+...+n^3)-6*(1^2+2^2+3^2+...+n^2)+4(1+2+3+4+...+n)-(n-1)-2 n^4-1=4*(1^3+2^3+3^3+...+n^3)-6*n(n+1)(2n+1)/6+4*n(n+1)/2-n-1 n^4-1=4*(1^3+2^3+3^3+...+n^3)-n(n+1)(2n+1)+2n(n+1)-n-1 n^4-1=4*(1^3+2^3+3^3+...+n^3)-n(n+1)(2n+1)+2n(n+1)-n-1 4*(1^3+2^3+3^3+...+n^3) =n^4-1+n(n+1)(2n+1)-2n(n+1)+n+1 =n^4-1+(n+1)(2n^2-n)+n+1 =n^4-1+(2n^3+n^2-n)+n+1 =n^4+2n^3+n^2 =(n^2+n)^2 =(n(n+1))^2 1^3+2^3+3^3+...+n^3 =[n(n+1)/2]^2

证明不等式：(1/n)^n+(2/n)^n+(3/n)^n+.+(n/n)^n [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 lim2^n +3^n/2^n+1+3^n+1 3(n-1)(n+3)-2(n-5)(n-2) lim(n+3)(4-n)/(n-1)(3-2n) lim(n^3+n)/(n^4-3n^2+1) n(n+1)(n+2)(n+3)+1 因式分解 n(n+1)(n+2)(n+3)+1等于多少 lim[(n+3)/(n+1))]^(n-2) 【n无穷大】 lim(2^n+3^n)^1 (n趋向无穷) 级数n/(n+1)(n+2)(n+3)和是多少 n*1+n*2+n*3+n*4.求公式判断n/(n+1)(n+2)(n+3)的收敛性