Amath 數學歸納法

Those part "LET, WHEN n=1, ASSUMPTION" are the same and I will just type once here:
Let P(n) be " ... ".
When n = 1, L.H.S. = ... = R.H.S.
So P(1) is true.
Assume that P(k) is true, i.e. "..."
1. When n = k + 1, L.H.S.
= (1)(2) + (2)(3) + (3)(4) + ... + (k)(k + 1) + (k + 1)(k + 2)
= k(k+1)(k+2) / 3 + (k + 1)(k + 2)
= (k + 1)(k + 2)(k/3 + 1)
= (1/3)(k + 1)(k + 2)(k + 3)
= R. H. S.
2. When n = k + 1, L.H.S.
= (1)(4) + (2)(7) + (3)(10) + ... + (k)(3k+1) +(k + 1)(3k + 4)
= k (k + 1)^2 + (k + 1)(3k + 4)
= (k + 1)(k^2 + 4k + 4)
= (k + 1)(k + 2)^2
= R. H. S.
3. When n = k + 1, L.H.S.
= 1^2/(1)(3) + 2^2/(3)(5) + ... + k^2/(2k - 1)(2k + 1) + (k + 1)^2/(2k + 1)(2k + 3)
= k (k + 1) / 2 (2k + 1) + (k + 1)^2 / (2k + 1)(2k + 3)
= (k + 1) / (2k + 1) [k/2 + (k + 1) / (2k + 3) ]
= (k + 1) / 2(2k + 1) [ (2k^2 + 3k + 2k + 2) / (2k + 3) ]
= (k + 1) / 2(2k + 1) [(2k + 1)(k + 2) / (2k + 3) ]
= (k + 1)(k + 2) / 2(2k + 3)
= R. H. S.
By Mathematical Induction, P(n) is true for all natural numbers n.