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.