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2. Bandar's Blunder (1 point) Bandar wants to use induction to prove the true statement that n! > n² for all n ? 4. His proof is incorrect, and it's your task to help him identify his mistake! Proof: • Inductive step: Let k be arbitrary. Assume k! > k². We need to show (k + 1)!> (k + 1)² (k + 1)!= (k + 1). k! > (k + 1). k² = (k + 1)(k \cdot k) ? (k+1)(2\cdot k) = (k + 1)(k+k) ? (k + 1)(k + 1) = (k + 1)² This proves (k + 1)! > (k + 1)². • Base Case: let n = 0, 0! = 1 > 0² = 0 Thus by mathematical induction, n! > n² for all n ? 0. What is wrong with Bandar's proof?

          2. Bandar's Blunder (1 point)
Bandar wants to use induction to prove the true statement that
n! > n² for all n ? 4.
His proof is incorrect, and it's your task to help him identify his mistake!
Proof:
• Inductive step:
Let k be arbitrary. Assume k! > k². We need to show (k + 1)!> (k + 1)²
(k + 1)!= (k + 1). k!
> (k + 1). k²
= (k + 1)(k \cdot k)
? (k+1)(2\cdot k)
= (k + 1)(k+k)
? (k + 1)(k + 1)
= (k + 1)²
This proves (k + 1)! > (k + 1)².
• Base Case: let n = 0, 0! = 1 > 0² = 0
Thus by mathematical induction, n! > n² for all n ? 0.
What is wrong with Bandar's proof?

2. Bandar's Blunder (1 point)
Bandar wants to use induction to prove the true statement that
n! > n² for all n ? 4.
His proof is incorrect, and it's your task to help him identify his mistake!
Proof:
• Inductive step:
Let k be arbitrary. Assume k! > k². We need to show (k + 1)!> (k + 1)²
(k + 1)!= (k + 1). k!
> (k + 1). k²
= (k + 1)(k ·k)
? (k+1)(2·k)
= (k + 1)(k+k)
? (k + 1)(k + 1)
= (k + 1)²
This proves (k + 1)! > (k + 1)².
• Base Case: let n = 0, 0! = 1 > 0² = 0
Thus by mathematical induction, n! > n² for all n ? 0.
What is wrong with Bandar's proof?

Added by Bego-A W.

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