First slide

Introduction to P.M.I

Question

For all $n \in N, 1 \times 1! + 2 \times 2! + 3 \times 3! + \dots + n \times n!$ is equal to

Easy

A

$(n + 1)! - 2$
B

$(n + 1)!$
C

$(n + 1)! - 1$
D

$(n + 1)! - 3$

Solution

Let the statement P(n) be defined as
$P (n) : 1 \times 1! + 2 \times 2! + 3 \times 3! \dots + n \times n! = (n + 1)! - 1$ for all natural numbers n.
for all natural numbers n.
$P (1) : 1 \times 1! = 1 = 2 - 1 = 2! - 1$
Assume that P(n) is true for some natural number k, i.e
$P (k) : 1 \times 1! + 2 \times 2! + 3 \times 3! + \dots . + k \times k! = (k + 1)! - 1$ ---------(i)
To prove P(k + 1) is true, we have
$\begin{matrix} P (k + 1) : 1 \times 1! + 2 \times 2! + 3 \times 3! + \dots + k \times k! + (k + 1) \times (k + 1)! \\ = (k + 1)! - 1 + (k + 1)! \times (k + 1) \\ = (k + 1 + 1) (k + 1)! - 1 \\ = (k + 2) (k + 1)! - 1 \\ = (k + 2)! - 1 \end{matrix}$
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.

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