mathematical induction?

In the question, the right-hand-side should be "-n^2(3+4n)" instead of "-n^2(3n+4n)".


Let P(n) : 1^3 - 2^3 + 3^3 - 4^3 + ... + (2n-1)^3 - (2n)^3 = -n^2(3+4n)
where n is a positive integer.

When n = 1 :
L.H.S = 1^3 - 2^3 = -7
R.H.S. = -1^2 × (3 + 4×1) = -7
L.H.S. = R.H.S.
Hence, P(1) is true.

Assume that P(k) is true, i.e. 1^3 - 2^3 + 3^3 - 4^3 + ... + (2k-1)^3 - (2k)^3 = -k^2(3+4k)
Prove that P(k+1) is true, i.e. 1^3 - 2^3 + 3^3 - 4^3 + ... + (2k+1)^3 - (2k+2)^3 = -(k+1)^2[3+4(k+1)]

Prove :
L.H.S.
= [1^3 - 2^3 + 3^3 - 4^3 + ... + (2k-1)^3 - (2k)^3] + (2k+1)^3 - (2k+2)^3
= -k^2(3+4k) + (8k^3 - 12k^2 + 6k + 1) - (8k^3 + 24k^2 + 24k + 8)
= -3k^2 - 4k^3 + 8k^3 + 12k^2 + 6k + 1 - 8k^3 - 24k^2 - 24k - 8
= -4k^3 -15k^2 - 18k - 7
= -[4k^3 +15k^2 + 18k + 7]
= -[(4k^3+7k^2) + (8k^2+14k) + (4k+7)]
= -[k^2(4k+7) + 2k(4k+7) + (4k+7)]
= -(k^2 + 2k + 1)(4k + 7)
= -(k+1)^2[3+4(k+1)]
= R.H.S.
Hence, when P(k) is true, P(k+1) must be true.

According the principle of mathematical induction, P(n) is true
where n is a positive integer.