1눀+2눀+3눀+....+n눀=?怎么证明？

在证明之间，先说明两个公式：

1. a³ - b³ = (a-b)(a²+ab+b²)

2. 1 + 2 + 3 + ……+n = n(n+1)/2

因为：

n³-(n-1)³ = [n-(n-1)][n²+n(n-1)+(n-1)²]

               = n² + n² - n + n² - 2n + 1

               = 3n² - 3n + 1

(n-1)³-(n-2)³ = 3(n-1)²-3(n-1)+1

(n-2)³-(n-3)³ = 3(n-2)²-3(n-2)+1

………………

4³     - 3³       = 3×4² - 3×4    +1

3³     - 2³       = 3×3² - 3×3    +1

2³     - 1³       = 3×2² - 3×2    +1

1³     - 0³       = 3×1² - 3×1    +1

把这些得到的式子左、右两边分别相加，可以得到：

n³ - 0³ = 3×(1²+2²+3²+……+n²) - 3×(1+2+3+……+n) + n×1

n³ + 3×(1+2+3+……+n) - n = 3×(1²+2²+3²+……+n²)

n³ + 3×n(n+1)/2 - n = 3×(1²+2²+3²+……+n²)

[2n³+3n(n+1)-2n]/2 = 3×(1²+2²+3²+……+n²)

[2n³+3n²+3n-2n]/2  = 3×(1²+2²+3²+……+n²)

(2n³+3n²+n)/2 = 3×(1²+2²+3²+……+n²)

n(2n²+3n+1)/2 = 3×(1²+2²+3²+……+n²)

n(n+1)(2n+1)/2 = 3×(1²+2²+3²+……+n²)

所以，

1²+2²+3²+……+n² = n(n+1)(2n+1)/6

1²+2²+3²+....+n²=n(n+1)(2n+1)/6

n(n+1)(2n+1)/6
用恒等式(1+n)³=1+3n+3n²+n³来证