Question about sum of integer involving factorial notation

[Sanosuke Sagara]

I have my question and my problem in the attachment that followed.

 

[quasar987]

Damn right you don't know how to continue. That is because you took r out of the sumation sign and you cannot do that because r is the thing that varies in your sumationg sign. r is nothing without the sumation sign!

The proof is easily done by induction.

 

[tacman]

After reviewing your attached question and problem, I can provide the following response:

Firstly, it is important to understand the concept of factorial notation in order to solve this problem. Factorial notation is represented by an exclamation mark (!) and is used to represent the product of all positive integers from 1 up to a given number. For example, 4! (read as "four factorial") is equal to 1 x 2 x 3 x 4 = 24.

Now, let's break down the given problem. We are asked to find the sum of all integers from 1 to n, where n is represented by the factorial notation (n!). In other words, we need to find the sum of 1 + 2 + 3 + ... + n.

To solve this problem, we can use the formula for the sum of an arithmetic sequence, which is Sn = (n/2)(a1 + an), where Sn is the sum, n is the number of terms, a1 is the first term, and an is the last term.

In our case, n represents the number of terms, which is equal to the number represented by the factorial notation (n!). The first term, a1, is 1, and the last term, an, is also n (since we are adding all integers from 1 to n).

Therefore, we can rewrite the formula as Sn = (n/2)(1 + n). Plugging in n! for n, we get Sn = (n!/2)(1 + n).

We also know that n! can be rewritten as n x (n-1) x (n-2) x ... x 2 x 1. So, we can further simplify the formula to Sn = (n x (n-1) x (n-2) x ... x 2 x 1)/2(1 + n).

This is the final formula for finding the sum of all integers from 1 to n, where n is represented by the factorial notation.

I hope this helps to clarify the problem and provide a solution. Please let me know if you have any further questions.