Mathematical Induction

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圖片參考：https://s.yimg.com/rk/HA07344914/o/1001065963.png

for all positive integers n.

Fact 1 (F1): 1*2 + 3*4 +...+ (2n-1)(2n) = [n(n+1)(4n-1)]/3
Fact 2 (F2): 1^2 +2^2 + 3^2 +...+ m^2 = [m(m+1)(2m+1)]/6
Identity: a^3 - b^3 = (a-b)(a^2 +ab+ b^2)
To prove: 1^3 - 2^3 + 3^3 - 4^3 +...+ (2n-1)^3 - (2n)^3 = -(n^2)(4n+3)

Using the Identity,
1^3 -2^3 = (1-2)(1^2 +1x2+ 2^2)
3^3 -4^3 = (3-4)(3^2 +3x4+ 4^2)
…
(2n-1)^3 –(2n)^3 = [(2n-1)-(2n)] [(2n-1)^2 +(2n-1)x(2n)+ (2n)^2]
Summing up on both sides,
LHS=1^3 - 2^3 + 3^3 - 4^3 +...+ (2n-1)^3 - (2n)^3
RHS= (1-2)(1^2 +1x2+ 2^2) + (3-4)(3^2 +3x4+ 4^2) + … + [(2n-1)-(2n)] [(2n-1)^2 +(2n-1)x(2n)+ (2n)^2]
= -{(1^2 +1x2+ 2^2) + (3^2 +3x4+ 4^2) + … + [(2n-1)^2 +(2n-1)x(2n)+ (2n)^2] }
= -{[(1^2 + 2^2) + (3^2 + 4^2) + … + [(2n-1)^2 + (2n)^2]] [1x2 +3x4 +… +(2n-1)x(2n)] }
= -{[1^2 + 2^2 + 3^2 + 4^2 + … + (2n-1)^2 + (2n)^2] [1x2 +3x4 +… +(2n-1)x(2n)] }
= -{[2n(2n+1)(4n+1)/6] + [n(n+1)(4n-1)]/3} by substituting F1 & F2
= -{n [(2n+1)(4n+1) + (n+1)(4n-1)]/3}
= -[n (8n^2+6n+1 + 4n^2+3n-1)/3]
= -[n (12n^2+9n)/3]
= -(n^2)(4n+3)
∴1^3 - 2^3 + 3^3 - 4^3 +...+ (2n-1)^3 - (2n)^3 = -(n^2)(4n+3) without using MI