mathematical induction唔識計呀

P(n): 2 + 5 + 8 + ... + (3n - 1) = n(3n + 1)/2
where n is natural number (1, 2, 3 …..)

When n = 1:
L.S. = 3(1) - 1 = 2
R.S. = 1[3(1) + 1]/2 = 2
Hence, P(1) is true.

Assume that n = k is true,
i.e. 2 + 5 + 8 + ... + (3k - 1) = k(3k + 1)/2
When n = k + 1:
Prove that: 2 + 5 + 8 + ... + [3(k + 1) - 1] = (k + 1)[3(k + 1) + 1]/2
L.S.
= 2 + 5 + 8 + ... + (3k - 1) + [3(k + 1) - 1]
= [2 + 5 + 8 + ... + (3k - 1)] + (3k + 2)
= [k(3k + 1)/2] + (3k + 2)
= [(3k² + k)/2] + [(6k + 4)/2]
= (3k² + 7k + 4)/2
= (k + 1)(3k + 4)/2
= (k + 1)(3k + 3 + 1)/2
= (k + 1) [3(k + 1) + 1]/2
= R.S.

By the principle of mathematical induction, P(n) is true where n is natural number.