超簡單數學歸納法2

Let P(n) be 2^2+4^2+6^2+...+(2n)^2=2/3n(n+1)(2n+1) .
P(1):
LHS:2^2=4
RHS:2/3(2)(3)=2/3(6)=12/3=4
P(1) is true.
Assume P(k) is true,P(k+1):
2^2+4^2+6^2+...+(2k)^2+[2(k+1)]^2
=2/3k(k+1)(2k+1)+[2(k+1)]^2
=(4k^3+6k^2+2k)/3+4k^2+8k+4
=4k^3/3+2k^2+4k^2+2k/3+8k+4
=4k^3/3+6k^2+26/3k+4
=(4k^3+18k^2+26k+12)/3
=(k+1)(4k^2+14k+12)/3
=(k+1)(k+2)(4k+6)/3
=2/3(k+1)(k+2)(2k+3)
=2/3(k+1)[(k+1)+1][2(k+1)+1]

PROOF IS COMPLETED

設P(n)為命題，

當n = 1，

L.H.S. = 2^2 = 4

R.H.S. = 2(1)[2 + 1](1 + 1) / 3 = 4 = L.H.S.

所以，P(1)成立。

假設P(k)成立，

當n = k + 1，

L.H.S. = 2(2k + 1)(k + 1) / 3 + (2k + 2)^2

= [2k(k + 1)(k + 1) + 12(k + 1)^2] / 3

= 2(k + 1)(2k^2 + 7k + 6) / 3

= 2(k + 1)(2k + 3)(k + 2) / 3

= R.H.S.

所以，P(k + 1)成立。

根據MI，對所有自然數，命題P(n)都成立。