³ ²

平方和公式

(n+1)² = (x+1)(x+1) = x(x+1) + (x+1) = x*x + x + x + 1 = x² + 2x +1
(n+1)² = (x-1)(x-1) = x(x-1) + -1(x-1) = x*x - x - x + 1 = x² + 2x +1

立方和公式

for: 

(n+1)² = n²+2n+1
(n-1)² = n²-2n+1

so:
(n+1)² - (n-1)² = (n²+2n+1) - (n²-2n+1)
(n+1)² - (n-1)² = 4n
n = [((n+1)² - (n-1)²]/4

then:

n³ = ¼n²[ -(n-1)² + (n+1)² ] = ¼[ -(n-1)²n² + n²(n+1)² ]


then:
1³+2³+3³+...+n³ = ¼(([ -(1-1)²1² 
    + 1²(1+1)²] + [ -(2-1)²2² 
	+ 2²(2+1)²] + [ -(3-1)²3²
	+ 3²(3+1)²]
	+ ...+[ -(n-1)²n² + n²(n+1)²] )
then:
1³+2³+3³+...+n³ = ¼n²(n+1)²

image

2 扩展=自然数和

1 + 2 + 3 + 4 ... + (n-1) + n = (1+n)+(2+n-1)+(3+n-2) + (4 + n -3) = n*[½(n+1)]

so:

(1+2+3+4+ ... + n)² = n*[½(n+1)]*n*[½(n+1)] = (½*½)n²(n+1)² = ¼n²(n+1)² = 1³+2³+3³+...+n³

3 扩展-立方和图解(归纳)法

image

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