請問有人可以幫我證明下面這個公式嗎 n Σ k^2 k二次=1/6 n(n+1)(2n+1) (k=1)?

以下 Σ 的計算範圍皆為 k = 1 到 k = n

claim : Σ k² = (1/6)n(n+1)(2n+1)
pf :
(k+1)³ = k³ + 3k² + 3k + 1
(k+1)³ - k³ = 3k² + 3k + 1
因此 :

(n+1)³ - n³ = 3n² + 3n + 1
n³ - (n-1)³ = 3(n-1)² + 3(n-1) + 1
(n-1)³ - (n-2)³ = 3(n-2)² + 3(n-2) + 1
...................................
4³ - 3³ = 3*3² + 3*3 + 1
3³ - 2³ = 3*2² + 3*2 + 1
2³ - 1³ = 3*1² + 3*1 + 1

以上 n 個式子加總得 :
(n+1)³ - 1³ = 3( Σ k² ) + 3( Σ k ) + n

Σ k²
= (1/3) * [ (n+1)³ - 1 - 3( Σ k ) - n ]
= (1/3) * [ (n+1)³ - 1 - 3(1/2)n(n+1) - n ]
= (1/3)(n+1) * [ (n+1)² - 1 - 3n/2 ]
= (1/6)(n+1) * [ 2(n+1)² - 2 - 3n ]
= (1/6)(n+1)( 2n² + n )
= (1/6)(n+1)n(2n+1)

Q.E.D.