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Q.
For all $n \in N, 1 \times 1 !+2 \times 2 !+3 \times 3 !+\cdots+n \times n !$ is equal to
Principle of Mathematical Induction
Solution:
Let the statement $P(n)$ be defined as
$P(n): 1 \times 1 !+2 \times 2 !+3 \times 3 ! \ldots+n \times n !=(n+1) !-1$ for all natural numbers $n$.
Note that $P(1)$ is true, since
$P(1): 1 \times 1 !=1=2-1=2 !-1$
Assume that $P(n)$ is true for some natural number $k$,
i.e., $P(k): 1 \times 1 !+2 \times 2 !+3 \times 3 !+\ldots+k \times k !=(k+1) !-1 ...$(i)
To prove $P(k+1)$ is true, we have
$P(k+1): 1 \times 1 ! +2 \times 2 !+3 \times 3 !+\cdots+k \times k ! +(k+1)\times(k+1)!$
$ =(k+1) !-1+(k+1) ! \times(k+1) \text { [by Eq. }(i)] $
$ =(k+1+1)(k+1) !-1 $
$ =(k+2)(k+1) !-1 $
$=(k+2) !-1$
Thus, $P(k+1)$ is true, whenever $P(k)$ is true. Therefore, by the principle of mathematical induction, $P(n)$ is true for all natural numbers $n$.