Question

The value of $1 + 1 \cdot 1! + 2 \cdot 2! + 3 \cdot 3! + \ldots \ldots + n \cdot n!$ is
A $\left( {n + 1} \right)! + 1$
B $\left( {n - 1} \right)! + 1$
C $\left( {n + 1} \right)! – 1$
D $\left( {n + 1} \right)!$

Answer 1

- Hint: In this problem, we first need to convert the given expression as a sum of $n + 1$ terms by adding and subtracting 1. Next, use the property of the factorial to simplify the obtained equation.

Complete step-by-step solution -
Consider the given expression $1 + 1 \cdot 1! + 2 \cdot 2! + 3 \cdot 3! + \ldots \ldots + n \cdot n!$ as$p$.
$P = 1 + 1 \cdot 1! + 2 \cdot 2! + 3 \cdot 3! + \ldots \ldots + n \cdot n!$

Now, rewrite the above expression as the sum of $n + 1$ terms as shown below.

$$\,\,\,\,\,\,P = 1 + \left( {2 - 1} \right)1! + \left( {3 - 1} \right)2! + \ldots \ldots + \left( {n + 1 - 1} \right)n! \\\ \Rightarrow P = 1 + 2 \cdot 1! - 1 \cdot 1! + 3 \cdot 2! - 1 \cdot 2! \ldots \ldots + \left( {n + 1} \right)n! - n! \\\ \Rightarrow P = 1 + 2! - 1! + 3! - 2! \ldots \ldots + \left( {n + 1} \right)! - n! \\\ \Rightarrow P = 1 + \left( {n + 1} \right)! - 1 \\\ \Rightarrow P = \left( {n + 1} \right)! \\\$$

Thus, the value of $1 + 1 \cdot 1! + 2 \cdot 2! + 3 \cdot 3! + \ldots \ldots + n \cdot n!$ is $\left( {n + 1} \right)!$, hence, the option (D) is the correct answer.

Note: The factorial of a number $n$ is a product of the natural numbers being $n$ along with$n$. For example: $5! = 5 \times 4 \times 3 \times 2 \times 1$. The factorial of a number cannot be negative. The factorial of 0 is 1.