SOLUTION: Find a closed form for S_n = 1! \cdot (1^2 + 1) + 2! \cdot (2^2 + 2) + \dots + n! \cdot (n^2 + n).\] for any integer n \ge 1. Your response should have a factorial.

Question 1209827: Find a closed form for
S_n = 1! \cdot (1^2 + 1) + 2! \cdot (2^2 + 2) + \dots + n! \cdot (n^2 + n).\]
for any integer n \ge 1. Your response should have a factorial.
Found 2 solutions by CPhill, ikleyn:
Answer by CPhill(1959)   (Show Source): You can put this solution on YOUR website!
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

Answer by ikleyn(52776)   (Show Source): You can put this solution on YOUR website!
.
Find a closed form for S_n = 1!*(1^2 + 1) + 2!*(2^2 + 2) + . . . + n!*(n^2 + n)
for any integer n >= 1
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

        The solution and the answer in the post by @CPhill both are .

                        Indeed,      let's check for n = 3.

Left side is

            1!*(1^2+1) + 2!*(2^2+2) + 3!*(3^2+3) = 1*(1+1) + 2*(2+2) + 6*(9+3) = 1*2 + 2*4 + 6*12 = 2 + 8 + 72 = 82.

Right side, according to @CPhill, is

            (3+2)! - 2 = 5! - 2 = 1*2*3*4*5 - 2 = 120 - 2 = 118.

But   82 =/= 118.

This is the , which ruins the solution by @CPhill to dust.

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                Regarding the post by @CPhill . . .

Keep in mind that @CPhill is a pseudonym for the Google artificial intelligence.

The artificial intelligence is like a baby now. It is in the experimental stage
of development and can make mistakes and produce nonsense without any embarrassment.

                It has no feeling of shame - it is shameless.

This time, again, it made an error.

Although the @CPhill' solutions are copy-paste Google AI solutions, there is one essential difference.

Every time, Google AI makes a note at the end of its solutions that Google AI is experimental
and can make errors/mistakes.

All @CPhill' solutions are copy-paste of Google AI solutions, with one difference:
@PChill never makes this notice and never says that his solutions are copy-past that of Google.
So, he NEVER SAYS TRUTH.

Every time, @CPhill embarrassed to tell the truth.
But I am not embarrassing to tell the truth, as it is my duty at this forum.

And the last my comment.

When you obtain such posts from @CPhill, remember, that NOBODY is responsible for their correctness,
until the specialists and experts will check and confirm their correctness.

Without it, their reliability is ZERO and their creadability is ZERO, too.

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