點用m.i prove 這條

Let P ( n ) be the proposition '1^3+3^3+5^3+...+(2n-1)^3 =n^2(2n^2 -1)'

When n = 1,

L.H.S. = ( 2 –1 )^3=1

R.H.S. = 1^2(2-1) = 1 = L.H.S.

So P ( 1 ) is true.

Assume P ( k ) is true for some positive integers k, i.e.

1^3+3^3+5^3+...+(2k-1)^3 =k^2(2k^2 -1)“

When n = k + 1,

L.H.S. =1^3+3^3+5^3+...+(2k-1)^3+(2k+1)^3

=k^2(2k^2 -1)+(2k+1)^3


= 2k^4-k^2+8k^3+12k^2+6k+1

=2k^4+8k^3+11k^2+6k+1

=(k+1)^2(2k^2+4k+1)

=(k+1)^2[2(k^2+2k)+1]

=(k+1)^2[2(k^2+2k+1-1)+1]

=(k+1)^2[2(k+1)^2+2-1]

R.H.S. =(k+1)^2[2(k+1)^2+2-1]

= L.H.S. So P ( k + 1 ) is true.

By the principle of mathematical induction, P ( n ) is true for all positive integers n.

Let P(n) be the proposition " 1^3+3^3+5^3+...+(2n-1)^3=n^2(2n^2-1) "
When n=1,
L.H.S.= (2*1-1)^3
=1
R.H.S = 1^2(2*1^2-1)
= 2-1
=1
Because L.H.S. = R.H.S.
Therefore, P(1) is true

Assume P(k) is true for some (positive integer/ natural number) k,
so that 1^3+3^3+5^3+...+(2k-1)^3=k^2(2k^2-1)

When n= k+1
L.H.S. = 1^3+3^3+5^3+...+(2k-1)^3 + [2(k+1)-1]^3
=k^2(2k^2-1)+ [2(k+1)-1]^3
=k^2(2k^2-1)+ (2k+1)^3
=k^2(2k^2-1)+ (8k^3+12k^2+6k+1)
=2k^4+k^2+8k^3+12k^2+6k+1
=2k^4+8k^3+13k^2+6k+1

R.H.S.=(k+1)^2[2(k+1)^2-1]
= (k^2+2K+1) (2k^2+4k+2-1)
=(k^2+2K+1)(2k^2+4k+1)
=2k^4+8k^3+13k^2+6k+1
Since L.H.S= R.H.S.
Therefore, P(k+1) is also true
By M.I., P(n) is true for all (positive integers/ natural numbers) n