Key Concepts and Formulas

• Telescoping Series: A series where consecutive terms cancel out, leaving only the first and last terms.
• Factorial Identity: $(r+1)r! = (r+1)!$
• Expressing Polynomials: Manipulating polynomials to fit a desired form, often involving $(r+1)$ or $(r+2)$ terms to interact with $r!$.

Step-by-Step Solution

Step 1: Problem Statement and General Term

We are given the sum: $S = \sum_{r=1}^{10} (r^2 + 1)r!$ Let the general term be $T_r = (r^2 + 1)r!$. Our goal is to express $T_r$ in the form $f(r+1) - f(r)$ for some function $f(r)$, which will allow the sum to telescope.

Step 2: Manipulating the General Term

We want to rewrite $r^2 + 1$ in a more useful form. We can express it as: $r^2 + 1 = r^2 + 2r + 1 - 2r = (r+1)^2 - 2r$ Now, substitute this back into $T_r$: $T_r = ((r+1)^2 - 2r)r! = (r+1)^2 r! - 2r \cdot r!$

Step 3: Applying Factorial Identities

Using the identity $(r+1)r! = (r+1)!$, we can rewrite the terms: $(r+1)^2 r! = (r+1)(r+1)!$ $2r \cdot r! = 2(r \cdot r!) = 2((r+1)! - r!)$ Substituting these into $T_r$: $T_r = (r+1)(r+1)! - 2((r+1)! - r!) = (r+1)(r+1)! - 2(r+1)! + 2r!$

Step 4: Combining Terms and Seeking Telescoping Form

Combine the terms involving $(r+1)!$: $T_r = (r+1 - 2)(r+1)! + 2r! = (r-1)(r+1)! + 2r!$ Now, let's try to write this as $f(r+1) - f(r)$. Let's consider $f(r) = (r-1)r!$. Then $f(r+1) = ((r+1) - 1)(r+1)! = r(r+1)!$ Therefore, $f(r+1) - f(r) = r(r+1)! - (r-1)r! = r(r+1)r! - (r-1)r! = (r^2 + r - r + 1)r! = (r^2 + 1)r! = T_r$ So, we have successfully expressed $T_r$ as $f(r+1) - f(r)$ where $f(r) = (r-1)r!$.

Step 5: Evaluating the Telescoping Sum

Now we evaluate the sum: $S = \sum_{r=1}^{10} (f(r+1) - f(r)) = (f(2) - f(1)) + (f(3) - f(2)) + \dots + (f(11) - f(10))$ This is a telescoping sum, so all intermediate terms cancel: $S = f(11) - f(1)$ Since $f(r) = (r-1)r!$, we have: $f(11) = (11-1)11! = 10 \cdot 11!$ $f(1) = (1-1)1! = 0 \cdot 1! = 0$ Thus, $S = 10 \cdot 11! - 0 = 10 \cdot 11!$

Step 6: Conclusion and Matching with Options

The sum evaluates to $10 \times (11!)$.

Comparing with the given options: (A) (11)! (B) 10 $\times$ (11!) (C) 101 $\times$ (10!) (D) 11 $\times$ (11!)

Our calculated sum, $10 \times (11!)$, matches option (B).

Common Mistakes & Tips

• Sign Errors: Be extremely careful with signs when manipulating the terms, especially when distributing negative signs.
• Incorrect Factorial Manipulation: Double-check your factorial identities. A common mistake is to incorrectly apply $(r+1)r! = (r+1)!$.
• Choosing the Right f(r): The choice of $f(r)$ is crucial. Practice with different forms to gain intuition. Starting with $(Ar+B)r!$ and solving for A and B can be a useful technique.

Summary

We evaluated the sum by expressing the general term as a difference of two terms, $f(r+1) - f(r)$, which resulted in a telescoping series. By carefully manipulating the polynomial part and using factorial identities, we found that $f(r) = (r-1)r!$. This allowed us to simplify the sum to $f(11) - f(1) = 10 \times (11!)$.

The final answer is \boxed{10 \times (11!)}, which corresponds to option (B).