The statement P(n): 1 times 1! + 2 times 2! + | vedclass

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(C) We check the statement $P(n)$ for $n = 1$: $LHS = 1 	imes 1! = 1$. $RHS = (1 + 1)! - 1 = 2! - 1 = 2 - 1 = 1$. Since $LHS = RHS$,the statement is

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True for all $n \in \mathbb{N}$

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(C) We check the statement $P(n)$ for $n = 1$: $LHS = 1 	imes 1! = 1$. $RHS = (1 + 1)! - 1 = 2! - 1 = 2 - 1 = 1$. Since $LHS = RHS$,the statement is true for $n = 1$. Assume the statement is true for $n = k$,i.e.,$1 	imes 1! + 2 	imes 2! + \dots + k 	imes k! = (k + 1)! - 1$. For $n = k + 1$,we have: $(1 	imes 1! + 2 	imes 2! + \dots + k 	imes k!) + (k + 1) 	imes (k + 1)! = ((k + 1)! - 1) + (k + 1) 	imes (k + 1)!$. $= (k + 1)! 	imes (1 + k + 1) - 1 = (k + 1)! 	imes (k + 2) - 1 = (k + 2)! - 1$. This matches the form $(n + 1)! - 1$ for $n = k + 1$. Thus,by the principle of mathematical induction,the statement is true for all $n \in \mathbb{N}$.

The statement $P(n): 1 \times 1! + 2 \times 2! + 3 \times 3! + \dots + n \times n! = (n + 1)! - 1$ is

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