在证明之间,先说明两个公式:
a³ - b³ = (a-b)(a²+ab+b²)
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