Rewritten Solution

1. Key Concept: Telescoping Sums and Factorial Manipulation

This problem requires us to evaluate a sum where each term is transformed into a difference of consecutive terms. This technique is known as a "telescoping sum," where intermediate terms cancel out, simplifying the sum to an expression involving only the first and last terms. The general form is: $\sum_{r=1}^{n} (A(r+1) - A(r)) = A(n+1) - A(1)$ or $\sum_{r=1}^{n} (A(r) - A(r-1)) = A(n) - A(0)$

A crucial property of factorials, $(n+1)! = (n+1) \cdot n!$, will be used extensively to manipulate the terms into the desired difference form. Specifically, we often use the identity $n \cdot n! = (n+1-1)n! = (n+1)n! - n! = (n+1)! - n!$.

2. Manipulating the General Term $(r^2+1)(r!)$.

Our primary goal is to rewrite the general term, $f(r) = (r^2+1)(r!)$, in a form that allows for telescoping summation. This typically involves expressing $f(r)$ as a difference involving factorials like $(g(r+1)! - g(r)!)$.

Let's start by expanding the term: $f(r) = (r^2+1)(r!) = r^2(r!) + r!$

Now, we focus on the $r^2(r!)$ term and strategically introduce higher-order factorials to facilitate cancellation later.

• Step 1: Transform $r \cdot r!$ using factorial properties. We can rewrite $r^2(r!)$ as $r \cdot (r \cdot r!)$. Applying the identity $r \cdot r! = (r+1)! - r!$: $r \cdot r! = (r+1-1)r! = (r+1)r! - r! = (r+1)! - r!$ Substituting this back into $r^2(r!)$: $r^2(r!) = r \cdot [(r+1)! - r!]$ $= r(r+1)! - r(r!)$

• Step 2: Combine with the original $+ r!$ term. Substitute this expanded form of $r^2(r!)$ back into the full expression for $f(r)$: $f(r) = [r(r+1)! - r(r!)] + r!$ $f(r) = r(r+1)! - r(r!) + r!$

• Step 3: Further transformation of terms to introduce $(r+2)!$. We need to rewrite $r(r+1)!$ to incorporate $(r+2)!$. We can express $r$ as $(r+2-2)$: $r(r+1)! = (r+2-2)(r+1)!$ $= (r+2)(r+1)! - 2(r+1)!$ Using the factorial property $(r+2)(r+1)! = (r+2)!$, we get: $r(r+1)! = (r+2)! - 2(r+1)!$

• Step 4: Substitute and simplify the remaining terms. Substitute this result back into the expression for $f(r)$: $f(r) = [(r+2)! - 2(r+1)!] - r(r!) + r!$ Now, let's simplify the terms $-r(r!) + r!$. We apply the identity $r \cdot r! = (r+1)! - r!$ again: $-r(r!) + r! = -[(r+1)! - r!] + r!$ $= -(r+1)! + r! + r!$ $= -(r+1)! + 2r!$

• Step 5: Assemble the final simplified expression for $f(r)$. Combine the results from Step 3 and Step 4: $f(r) = (r+2)! - 2(r+1)! - (r+1)! + 2r!$ $f(r) = (r+2)! - 3(r+1)! + 2r!$

3. Expressing $f(r)$ as a Difference for Telescoping

To make the summation telescope, we group the terms in $f(r)$ into differences of the form $g(k) - g(k-1)$. Notice that we have $(r+2)!$, $(r+1)!$, and $r!$. Let's group them as follows: $f(r) = [(r+2)! - (r+1)!] - 2[(r+1)! - r!]$

Explanation:

• The term $[(r+2)! - (r+1)!]$ is a difference where the first term starts one step higher than the second. This will lead to cancellations. If we define $g(k) = k!$, this part is $g(r+2) - g(r+1)$. Alternatively, if $G(r) = (r+1)!$, this is $G(r+1) - G(r)$.
• Similarly, $[(r+1)! - r!]$ is a difference where the first term is one step higher than the second. If we define $H(r) = r!$, this is $H(r+1) - H(r)$.

This specific grouping is key to using the telescoping sum property effectively.

4. Evaluating the Summation

Now we can apply the summation from $r=1$ to $20$ to our transformed $f(r)$: $\sum_{r=1}^{20} f(r) = \sum_{r=1}^{20} \left( [(r+2)! - (r+1)!] - 2[(r+1)! - r!] \right)$ We can split this into two separate telescoping sums: $S = \sum_{r=1}^{20} [(r+2)! - (r+1)!] \quad - \quad 2 \sum_{r=1}^{20} [(r+1)! - r!]$

• Evaluating the First Sum: $\sum_{r=1}^{20} [(r+2)! - (r+1)!]$ Let $G(r) = (r+1)!$. The sum is of the form $\sum_{r=1}^{20} [G(r+1) - G(r)]$. Using the telescoping sum property, this sum evaluates to $G(20+1) - G(1)$: $= G(21) - G(1) = (21+1)! - (1+1)! = 22! - 2! = 22! - 2$ (To illustrate, writing out terms: $(3!-2!) + (4!-3!) + \dots + (22!-21!) = 22! - 2!$)

• Evaluating the Second Sum: $\sum_{r=1}^{20} [(r+1)! - r!]$ Let $H(r) = r!$. The sum is of the form $\sum_{r=1}^{20} [H(r+1) - H(r)]$. Using the telescoping sum property, this sum evaluates to $H(20+1) - H(1)$: $= H(21) - H(1) = 21! - 1! = 21! - 1$ (To illustrate, writing out terms: $(2!-1!) + (3!-2!) + \dots + (21!-20!) = 21! - 1!$)

• Combining the Results: Now substitute these results back into the total sum $S$: $S = (22! - 2) - 2(21! - 1)$ $S = 22! - 2 - 2 \cdot 21! + 2 \cdot 1$ $S = 22! - 2 - 2 \cdot 21! + 2$ $S = 22! - 2 \cdot 21!$

• Step 6: Final Simplification To match the options, we can factor out $21!$ from the expression. Recall that $22! = 22 \cdot 21!$. $S = 22 \cdot 21! - 2 \cdot 21!$ $S = (22 - 2) \cdot 21!$ $S = 20 \cdot 21!$

5. Verification with Options

Let's check which option matches our result $20 \cdot 21!$:

• (A) $22! - 21! = 22 \cdot 21! - 21! = (22-1)21! = 21 \cdot 21!$
• (B) $22! - 2(21!) = 22 \cdot 21! - 2 \cdot 21! = (22-2)21! = 20 \cdot 21!$
• (C) $21! - 2(20!) = 21 \cdot 20! - 2 \cdot 20! = (21-2)20! = 19 \cdot 20!$
• (D) $21! - 20! = 21 \cdot 20! - 20! = (21-1)20! = 20 \cdot 20!$

Our calculated sum, $20 \cdot 21!$, matches option (B).

Important Note: The problem statement indicates that option (A) is the correct answer. However, based on the standard mathematical derivation and common techniques for telescoping sums involving factorials, the result consistently points to option (B). It's possible there is a typo in the question's stated "Correct Answer" or a different identity/approach was intended. Assuming the problem and options are given correctly, the provided solution and this elaborated solution lead to (B).

6. Tips and Common Mistakes

• Don't Rush Simplification: Break down complex terms like $(r^2+1)(r!)$ step-by-step. Trying to find the final telescoping form directly can be difficult.
• Utilize Factorial Properties: The identity $n \cdot n! = (n+1)! - n!$ is extremely powerful for these types of sums. Remember it!
• Verify Your Transformation: After rewriting the general term $f(r)$ into a difference form, always expand it back to ensure it matches the original term. This catches algebraic errors early.
• Careful with Limits: When evaluating telescoping sums, ensure you correctly identify the terms that remain after cancellation (the first and last terms based on the sum's lower and upper limits).
• Double-Check Arithmetic: Even simple additions/subtractions with factorials can be error-prone.

7. Summary and Key Takeaway

This problem demonstrates a classic technique for evaluating sums involving factorials by transforming each term into a difference, thus creating a telescoping series. The key is the algebraic manipulation of $(r^2+1)(r!)$ into the form $[(r+2)! - (r+1)!] - 2[(r+1)! - r!]$. This allows for straightforward cancellation across the summation, yielding the final result. Mastering factorial identities and the concept of telescoping sums is vital for solving such problems efficiently in competitive exams like JEE.