Evaluate the Summation sum from x=1 to 4 of (x+1)!
You are asked to calculate the total value obtained when you sum the results of a mathematical operation that involves factorials for a series of consecutive integer values, starting from 1 and ending at 4. The operation you're summing over consists of taking each integer 'x', adding 1 to it, and then calculating the factorial of the result, which is denoted by (x+1)!. You'd perform this operation for every integer from 1 to 4 and then add all those computed factorial values together to get the final summation total.
$\sum_{x = 1}^{4} \left(\right. x + 1 \left.\right) !$
Answer
Solution:
Step 1:
Write out the summation for each integer value of $x$ from $1$ to $4$: $(1+1)! + (2+1)! + (3+1)! + (4+1)!$.
Step 2:
Calculate and simplify the factorial expressions for each term in the summation.
Step 2.1:
Begin simplifying each factorial term individually.
Step 2.1.1:
Combine $1$ and $1$ to get $(2)! + (2+1)! + (3+1)! + (4+1)!$.
Step 2.1.2:
Calculate $(2)!$ as $2 \times 1$: $2 \times 1 + (2+1)! + (3+1)! + (4+1)!$.
Step 2.1.3:
Perform the multiplication $2 \times 1$: $2 + (2+1)! + (3+1)! + (4+1)!$.
Step 2.1.4:
Combine $2$ and $1$ to get $2 + (3)! + (3+1)! + (4+1)!$.
Step 2.1.5:
Calculate $(3)!$ as $3 \times 2 \times 1$: $2 + 3 \times 2 \times 1 + (3+1)! + (4+1)!$.
Step 2.1.6:
Complete the multiplication for $3 \times 2 \times 1$.
Step 2.1.6.1:
Multiply $3$ by $2$: $2 + 6 \times 1 + (3+1)! + (4+1)!$.
Step 2.1.6.2:
Multiply $6$ by $1$: $2 + 6 + (3+1)! + (4+1)!$.
Step 2.1.7:
Combine $3$ and $1$ to get $2 + 6 + (4)! + (4+1)!$.
Step 2.1.8:
Calculate $(4)!$ as $4 \times 3 \times 2 \times 1$: $2 + 6 + 4 \times 3 \times 2 \times 1 + (4+1)!$.
Step 2.1.9:
Complete the multiplication for $4 \times 3 \times 2 \times 1$.
Step 2.1.9.1:
Multiply $4$ by $3$: $2 + 6 + 12 \times 2 \times 1 + (4+1)!$.
Step 2.1.9.2:
Multiply $12$ by $2$: $2 + 6 + 24 \times 1 + (4+1)!$.
Step 2.1.9.3:
Multiply $24$ by $1$: $2 + 6 + 24 + (4+1)!$.
Step 2.1.10:
Combine $4$ and $1$ to get $2 + 6 + 24 + (5)!$.
Step 2.1.11:
Calculate $(5)!$ as $5 \times 4 \times 3 \times 2 \times 1$: $2 + 6 + 24 + 5 \times 4 \times 3 \times 2 \times 1$.
Step 2.1.12:
Complete the multiplication for $5 \times 4 \times 3 \times 2 \times 1$.
Step 2.1.12.1:
Multiply $5$ by $4$: $2 + 6 + 24 + 20 \times 3 \times 2 \times 1$.
Step 2.1.12.2:
Multiply $20$ by $3$: $2 + 6 + 24 + 60 \times 2 \times 1$.
Step 2.1.12.3:
Multiply $60$ by $2$: $2 + 6 + 24 + 120 \times 1$.
Step 2.1.12.4:
Multiply $120$ by $1$: $2 + 6 + 24 + 120$.
Step 2.2:
Add the numbers together to simplify the expression.
Step 2.2.1:
Combine $2$ and $6$: $8 + 24 + 120$.
Step 2.2.2:
Add $8$ to $24$: $32 + 120$.
Step 2.2.3:
Combine $32$ and $120$ to get the final result: $152$.
Knowledge Notes:
The problem involves evaluating a summation of factorials. A factorial, denoted by an exclamation mark (!), is the product of all positive integers less than or equal to a given positive integer. For example, $5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$.
Key points to remember:
The factorial of a number $n$ is represented as $n!$ and is the product of all positive integers from $1$ to $n$.
The factorial of $0$ is defined as $1$, i.e., $0! = 1$.
Factorials are used in permutations and combinations, as well as in series and probability.
When evaluating a series involving factorials, it is important to calculate each term's factorial separately and then sum them up.
Simplification of factorials involves performing the multiplication of the sequence of descending natural numbers until reaching $1$.
The summation symbol $\Sigma$ represents the sum of a sequence of terms, which can be simplified by calculating each term and then adding them together.
