mathematical induction

Let P(n) be the proposition that 1*4 + 2*7 + 3*10 + ... + n(3n + 1) = n(n + 1)^2 for all natural numbers n
P(1) is true since LHS = 1*4 = 4
RHS = (1)(1 + 1)^2 = 4
Assume P(k) is true, i.e.
1*4 + 2*7 + 3*10 + ... + k(3k + 1) = k(k + 1)^2
then
1*4 + 2*7 + 3*10 + ... + k(3k + 1) + (k + 1)[3(k + 1) + 1]
= k(k + 1)^2 + (k + 1)[3(k + 1) + 1]
= (k + 1)[k(k + 1) + 3(k + 1) + 1]
= (k + 1)[k^2 + k + 3k + 4]
= (k + 1)[k^2 + 4k + 4]
= (k + 1)(k + 2)^2
= (k + 1)[(k + 1) + 1]^2
=> P(k + 1) is also true.
Hence by the principle of mathematical induction, P(n) is true for all natrual numbers.
Let P(n) be the proposition that 1＋4＋4^2 + …＋4^(n - 1) = (1/3)[(4^n) - 1] for all natural numbers n.
P(1) is true since LHS = 1
RHS = (1/3)(4 - 1) = 1
Assume P(k) is true, i.e.
1＋4＋4^2 + …＋4^(k - 1) = (1/3)[(4^k) - 1]
then
1＋4＋4^2 + …＋4^(k - 1) + 4^k
= (1/3)[(4^k) - 1] + 4^k
= (1/3)[(4^k) - 1 + (3)4^k]
= (1/3)[(4^k) - 1 + (4 - 1)4^k]
= (1/3)[(4^k) - 1 + (4)4^k - 4^k]
= (1/3)[(4)4^k - 1]
= (1/3)[4^(k+1) - 1]
=> P(k + 1) is also true.
Hence by the principle of mathematical induction, P(n) is true for all natrual numbers.