If `sum_(t=1)^(1003) (r^(2) + 1)r! = a! - b(c!)` where `a, b, c in N` the least value of `(a + b + c)` is `pqrs` then

Options

A. `p + q+ r + s = 4`
B. `(p + q)/(r + s) = 1`
C. `(p + q + r)/(s) = 3`
D. `p.q.r.s` is even

Explanation

To solve the problem, we need to evaluate the summation and express it in the form \( a! - b(c!) \).

### Step 1: Write the summation
We start with the given summation:
\[ S = \sum_{r=1}^{1003} (r^2 + 1) r! \]

### Step 2: Simplify the expression
We can rewrite \( r^2 + 1 \) as:
\[ r^2 + 1 = r^2 + r - r + 1 = (r^2 + r) - (r - 1) \]
Thus, we can express the summation as:
\[ S = \sum_{r=1}^{1003} \left( (r^2 + r) r! - (r - 1) r! \right) \]

### Step 3: Factor out \( r! \)
Now, we can factor out \( r! \):
\[ S = \sum_{r=1}^{1003} \left( r(r + 1)! - (r - 1) r! \right) \]

### Step 4: Rewrite the terms
The term \( r(r + 1)! \) can be rewritten as:
\[ r(r + 1)! = (r + 1)! - (r - 1) r! \]
So, we have:
\[ S = \sum_{r=1}^{1003} \left( (r + 1)! - (r - 1) r! \right) \]

### Step 5: Evaluate the summation
Now we can evaluate the summation:
\[ S = \sum_{r=1}^{1003} (r + 1)! - \sum_{r=1}^{1003} (r - 1) r! \]
The first term simplifies to:
\[ \sum_{r=1}^{1003} (r + 1)! = 1004! - 1! \]
The second term simplifies to:
\[ \sum_{r=1}^{1003} (r - 1) r! = \sum_{r=2}^{1003} (r - 1) r! = 1003! - 0! \]

### Step 6: Combine the results
Combining these results, we have:
\[ S = 1004! - 1 - (1003! - 1) = 1004! - 1003! \]

### Step 7: Express in the required form
We can express this as:
\[ S = 1005 \cdot 1004! - 2 \cdot 1004! \]
Thus, we can write:
\[ S = 1005 \cdot 1004! - 2 \cdot 1004! \]

### Step 8: Identify \( a, b, c \)
From the expression \( S = a! - b(c!) \), we can identify:
- \( a = 1005 \)
- \( b = 2 \)
- \( c = 1004 \)

### Step 9: Calculate \( a + b + c \)
Now, we calculate:
\[ a + b + c = 1005 + 2 + 1004 = 2011 \]

### Final Result
Thus, the least value of \( (a + b + c) \) is \( 2011 \).