Mathematical Induction ABC Questions?

a.)
P(n): 1² + 2² + ... + n² = n(n+1)(2n+1)/6 for positive integer n.

When n = 1:
L.H.S. = 1 and R.H.S. = 1
Then, L.H.S. = R.H.S.
Hence, P(1) is true.

Assumming that P(k) is true, i.e. 1² + 2² + ... + k² = k(k+1)(2k+1)/6
Prove that P(k+1) is true, i.e. 1² + 2² + ... + (k+1)² = (k+1)(k+2)(2k+3)/6

Proof:
L.H.S.
= (1² + 2² + ... + k²) + (k+1)²
= [k(k+1)(2k+1)/6] + [6(k+1)²/6]
= [(k+1)/6] [k(2k+1) + 6(k+1)]
= [(k+1)/6] (2k²+ k + 6k + 6)
= [(k+1)/6] (2k²+ 7k + 6)
= [(k+1)/6] (k+2)(2k+3)
= (k+1)(k+2)(2k+3)/6
= R.H.S.

P(1) is true.
Assuming P(k) is true, P(k+1) is also true.

By the principle of Mathematical Induction, P(n) is true for the positive integer n.

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b.)
P(n): 1³ + 2³ + ... + n³ = [n(n+1)/2]² for positive integer n.

When n = 1:
L.H.S. = 1 and R.H.S. = 1
Then, L.H.S. = R.H.S.
Hence, P(1) is true.

Assumming that P(k) is true, i.e. 1³ + 2³ + ... + k³ = [k(k+1)/2]²
Prove that P(k+1) is true, i.e. 1³ + 2³ + ... + (k+1) ³ = [(k+1)(k+2)/2]²

Proof:
L.H.S.
= 1³ + 2³ + ... + k³ + (k+1)³
= [k(k+1)/2]² + (k + 1)³
= [k²(k+1)²/4] + [4(k + 1)³/4]
= [(k+1)²/2²] [k² + 4(k + 1)]
= [(k+1)²/2²] (k² + 4k + 4)²
= [(k+1)²/2²] (k+2)²
= [(k+1)(k+2)/2]²
= R.H.S.

P(1) is true.
Assuming P(k) is true, P(k+1) is also true.

By the principle of Mathematical Induction, P(n) is true for the positive integer n.

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c.)
P(n): 2n < n! where in n > 4

When n = 5:
L.H.S. = 10 and R.H.S. = 120
Then, L.H.S. < R.H.S.
Hence, P(5) is true.

Assumming that P(k) is true, i.e. 2k - k! < 0
Prove that P(k+1) is true, i.e. 2(k+1) - (k+1)!

Proof:
2(k+1) - (k+1)!
= 2k + 2 - (k+1)! + k! - k!
= (2k - k!) + [k! - (k+1)!] + 2
= (2k - k!) + [k! - (k+1)•k!] + 2
= (2k - k!) + k![1 - (k+1)] + 2
= (2k - k!) + (2 - k•k!) < 0
for 2k - k! < 0 and 2 - k•k! < 0 when k > 4

P(5) is true.
Assuming P(k) is true, P(k+1) is also true.

By the principle of Mathematical Induction, P(n) is true for n > 4.