Question

Find $\mathbf{\sum }_{\mathbf{j}\mathbf{=}\mathbf{0}}^{\mathbf{4}}\mathbf{ }\mathbf{j}\mathbf{!}$

Short answer

The value of summation is .

Step-by-step solution

Determination of factorial of numbers

The product sigma gives the multiplication of all the numbers from lower limit to upper limit.

The factorial of 0 is given as:

0! = 1

The factorial of 1 is given as:

1! - (1) (1-1)

1! = 0

The factorial of 2 is given as:

2! = (2)(2-1)

2! = 2

The factorial of 3 is given as:

3! = (3) (3-1)(3-2)

3! = 6

The factorial of 4 is given as:

4! = (4)(4-1)(4-2)(4-3)

4! = 24

Determination of value of summation

The value of the product is given as:

$\begin{array}{r}\sum _{j=0}^4 j!=(0!)+(1!)+(2!)+(3!)+(4!)\\ \sum _{j=0}^4 j!=1+0+2+6+24\\ \sum _{j=0}^4 j!=33\end{array}$

Therefore, the value of sum is 33 .