If `S_(n)=(1^(2)-1+1)(1!)+(2^(2)-2+1)(2!)+...+(n^(2)-n+1)(n!)`, then `S_(50)=`

Options

A. `52!`
B. `1+49xx5!`
C. `52!-1`
D. `50xx51!-1`

Explanation

To solve for \( S_{50} \) in the given series

\[ S_n = (1^2 - 1 + 1)(1!) + (2^2 - 2 + 1)(2!) + \ldots + (n^2 - n + 1)(n!) \]

we can break down the expression step by step.

### Step 1: Simplify the General Term
The general term of the series can be expressed as:

\[ t_n = (n^2 - n + 1)(n!) \]

### Step 2: Rewrite the General Term
We can rewrite \( n^2 - n + 1 \) as follows:

\[ n^2 - n + 1 = n(n - 1) + 1 \]

Thus, we can express \( t_n \) as:

\[ t_n = (n(n - 1) + 1)(n!) = n(n - 1)(n!) + (n!) \]

### Step 3: Factor Out \( n! \)
Now, we can factor \( n! \) out of the expression:

\[ t_n = n(n - 1)(n!) + (n!) = n(n - 1)(n!) + 1(n!) \]

### Step 4: Write the Series
Now we can write the series \( S_n \):

\[ S_n = \sum_{k=1}^{n} t_k = \sum_{k=1}^{n} (k(k - 1)(k!) + k!) \]

This can be split into two separate sums:

\[ S_n = \sum_{k=1}^{n} k(k - 1)(k!) + \sum_{k=1}^{n} k! \]

### Step 5: Evaluate Each Sum
1. The first sum \( \sum_{k=1}^{n} k(k - 1)(k!) \) can be simplified using the identity \( k(k - 1)(k!) = (k + 1)! - k! \):

\[ \sum_{k=1}^{n} k(k - 1)(k!) = \sum_{k=1}^{n} ((k + 1)! - k!) = (2! - 1!) + (3! - 2!) + \ldots + ((n + 1)! - n!) \]

This telescopes to:

\[ (n + 1)! - 1! \]

2. The second sum \( \sum_{k=1}^{n} k! \) is simply the sum of factorials from \( 1! \) to \( n! \).

### Step 6: Combine the Results
Thus, we have:

\[ S_n = (n + 1)! - 1 + \sum_{k=1}^{n} k! \]

### Step 7: Calculate \( S_{50} \)
Now, substituting \( n = 50 \):

\[ S_{50} = (51)! - 1 + \sum_{k=1}^{50} k! \]

### Step 8: Evaluate the Sum of Factorials
The sum \( \sum_{k=1}^{50} k! \) is a large number, but we can denote it as \( C \) for simplicity. Thus:

\[ S_{50} = 51! - 1 + C \]

### Final Result
Since the problem asks for \( S_{50} \) and we have simplified it down, we can conclude that:

\[ S_{50} = 51! - 1 \]

### Answer
Thus, the answer is:

\[ \boxed{51! - 1} \]