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Question 1209826: For a positive integer k, let
S_k = 1 \cdot 1! \cdot 2 + 2 \cdot 2! \cdot 3 + \dots + k \cdot k! \cdot (k + 1).
Find a closed form for S_k.
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_k.
**1. Analyze the Terms**
The general term of the sum is:
T_n = n * n! * (n + 1)
**2. Manipulate the Terms**
We can rewrite the term as:
T_n = n * n! * (n + 1) = n * (n + 1)!
Now, we can rewrite n as (n + 2 - 2):
T_n = (n + 2 - 2) * (n + 1)!
Distribute:
T_n = (n + 2) * (n + 1)! - 2 * (n + 1)!
T_n = (n + 2)! - 2 * (n + 1)!
**3. Apply the Summation**
S_k = Σ T_n (from n = 1 to k)
S_k = Σ [(n + 2)! - 2 * (n + 1)!] (from n = 1 to k)
S_k = [3! - 2 * 2!] + [4! - 2 * 3!] + [5! - 2 * 4!] + ... + [(k + 2)! - 2 * (k + 1)!]
**4. Observe the Pattern**
Notice that we can rearrange the terms:
S_k = [3! + 4! + 5! + ... + (k + 2)!] - 2 * [2! + 3! + 4! + ... + (k + 1)!]
Let's look at the first few terms:
* 3! - 2 * 2! = 6 - 4 = 2
* 4! - 2 * 3! = 24 - 12 = 12
* 5! - 2 * 4! = 120 - 48 = 72
**5. Simplify the Sum**
We can rewrite T_n as:
T_n = (n + 2)! - 2 * (n + 1)! = [(n + 2) * (n + 1)!] - 2 * (n + 1)! = (n + 2 - 2) * (n + 1)! = n * (n + 1)!
Let's try a different approach.
T_n = n * n! * (n + 1) = n * (n + 1)!
Now, we can write:
(n + 1)! = (n + 1) * n!
T_n = n * (n + 1) * n!
We can also write:
(n + 1) * n! = (n + 1)!
T_n = n * (n + 1)!
We can rewrite n as (n + 2 - 2):
T_n = (n + 2 - 2) * (n + 1)! = (n + 2)! - 2(n + 1)!
Now, let's look at the partial sums:
* S_1 = 1 * 1! * 2 = 2
* S_2 = 1 * 1! * 2 + 2 * 2! * 3 = 2 + 12 = 14
* S_3 = 14 + 3 * 3! * 4 = 14 + 72 = 86
Let's try a different manipulation:
T_n = n * n! * (n + 1) = [(n + 2) - 2] * n! * (n + 1) = (n + 2)(n + 1)n! - 2(n + 1)n! = (n + 2)! - 2(n + 1)!
Now, let's sum:
S_k = Σ [(n + 2)! - 2(n + 1)!]
S_k = (3! - 2 * 2!) + (4! - 2 * 3!) + (5! - 2 * 4!) + ... + ((k + 2)! - 2(k + 1)!)
S_k = (k + 2)! - 2(k + 1)! - (2 * 2! - 3!) = (k+2)! - 2(k+1)! - (4 - 6) = (k+2)! - 2(k+1)! + 2
S_k = (k + 2)(k + 1)! - 2(k + 1)! + 2 = (k + 2 - 2)(k + 1)! + 2 = k(k + 1)! + 2
**6. Verify the Closed Form**
* S_1 = 1(2!) + 2 = 2 + 2 = 4 (incorrect)
* S_2 = 2(3!) + 2 = 12 + 2 = 14 (correct)
* S_3 = 3(4!) + 2 = 72 + 2 = 74 (incorrect)
The given closed form is not correct. We must find another form.
T_n = n * n! * (n + 1) = n * (n + 1)! = (n + 1 - 1)(n + 1)! = (n + 1)(n + 1)! - (n + 1)! = (n + 2 - 1)(n+1)!-(n+1)!
T_n = (n+2)!/(n+2) * (n+1) - (n+1)! = (n+2-1)(n+1)!-(n+1)! = (n+2)!(1-1/(n+2)) - (n+1)!
T_n = (n+2)! - (n+1)!-(n+1)! = (n+2)!-(n+1)!-(n+1)!
T_n = (n+2)! - 2(n+1)!
S_k = Σ [(n+2)!-2(n+1)!] = (k+2)! - 2(k+1)! - (3! - 2(2!)) = (k+2)! - 2(k+1)! - (6-4) = (k+2)! - 2(k+1)! - 2
S_k = (k+2)(k+1)! - 2(k+1)! - 2 = (k+2-2)(k+1)! - 2 = k(k+1)! - 2
**Final Answer:** S_k = k(k + 1)! - 2
Answer by ikleyn(52776)  (Show Source):
You can put this solution on YOUR website! .
S_k = 1*(1^2) + 2!*(2^2)*3 + . . . + k*k!*(k + 1).
Find a closed form for S_k.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
The solution and the answer in the post by @CPhill both are  .
Indeed, let's check for k = 3.
Left side is
1!*(1^2) + 2!*(2^2)*3 + 3*3!*(3+1) = 1*(1) + 2*(4)*3 + 3*(6)*4 = 1 + 24 + 72 = 97.
Right side, according to @CPhill, is
3*(3+1)! - 2 = 3*4! - 2 = 3*24 - 2 = 72 - 2 = 70.
But 97 =/= 70.
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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