Question 1209827
Let's find a closed form for the sum S_n.

**1. Analyze the General Term**

The general term of the sum is:

T_k = k! * (k² + k)

**2. Simplify the General Term**

We can factor out k from the parentheses:

T_k = k! * k(k + 1)

Rearrange the terms:

T_k = k * (k + 1) * k!

Notice that (k + 1) * k! = (k + 1)!

So, T_k = k * (k + 1)!

**3. Rewrite k**

We can rewrite k as (k + 2 - 2):

T_k = (k + 2 - 2) * (k + 1)!

Distribute:

T_k = (k + 2) * (k + 1)! - 2 * (k + 1)!

Recognize that (k + 2) * (k + 1)! = (k + 2)!

So, T_k = (k + 2)! - 2 * (k + 1)!

**4. Apply the Summation**

S_n = Σ[k=1 to n] T_k

S_n = Σ[k=1 to n] [(k + 2)! - 2 * (k + 1)!]

S_n = [3! - 2 * 2!] + [4! - 2 * 3!] + [5! - 2 * 4!] + ... + [(n + 2)! - 2 * (n + 1)!]

**5. Observe the Telescoping Pattern**

Notice that many terms cancel out:

S_n = [3! - 2 * 2!] + [4! - 2 * 3!] + [5! - 2 * 4!] + ... + [(n + 2)! - 2 * (n + 1)!]

S_n = -2 * 2! + (3! - 2 * 3!) + (4! - 2 * 4!) + ... + (n + 2)!

S_n = -2 * 2! -3! -4! ... +(n+2)!

S_n = (n + 2)! - (2(2!) + 3! + 4! + ... + (n+1)!)

Now let's examine a different approach.

S_n = Σ[k=1 to n] k * (k + 1)!

We can rewrite k as (k + 1 - 1):

S_n = Σ[k=1 to n] (k + 1 - 1) * (k + 1)!

S_n = Σ[k=1 to n] [(k + 1) * (k + 1)! - (k + 1)!]

S_n = Σ[k=1 to n] [(k + 2)! - (k + 1)!]

This is a telescoping sum:

S_n = [3! - 2!] + [4! - 3!] + [5! - 4!] + ... + [(n + 2)! - (n + 1)!]

S_n = (n + 2)! - 2!

S_n = (n + 2)! - 2

**Final Answer:**

S_n = (n + 2)! - 2