Understanding the Problem We are asked to find the value of the sum $\sum\limits_{r = 0}^{20} {{r^2}.{}^{20}{C_r}}$, where ${{}^{20}{C_r}}$ represents the binomial coefficient in the expansion of $(1+x)^{20}$. This type of problem often involves using properties of binomial coefficients or differentiation of binomial expansions.

Key Concepts and Formulas

1. Binomial Theorem: The expansion of $(1+x)^n$ is given by $\sum_{r=0}^n {{n}{C_r}} x^r$.
2. Sum of Binomial Coefficients: $\sum_{r=0}^n {{n}{C_r}} = 2^n$. This is obtained by setting $x=1$ in the binomial expansion.
3. Important Identities for Binomial Coefficients:
   • $r \cdot {{n}{C_r}} = n \cdot {{n-1}{C_{r-1}}}$ for $r \geq 1$. (When $r=0$, $r \cdot {{n}{C_r}} = 0$)
   • $r(r-1) \cdot {{n}{C_r}} = n(n-1) \cdot {{n-2}{C_{r-2}}}$ for $r \geq 2$. (When $r=0, 1$, $r(r-1) \cdot {{n}{C_r}} = 0$)

These identities are crucial for simplifying sums involving products of $r$ or $r(r-1)$ with binomial coefficients.

Step-by-Step Solution

Our goal is to evaluate $S = \sum\limits_{r = 0}^{20} {{r^2}.{}^{20}{C_r}}$.

Step 1: Decompose $r^2$ We can rewrite $r^2$ as $r(r-1) + r$. This decomposition is strategic because it allows us to use the two binomial identities mentioned above. $S = \sum_{r=0}^{20} (r(r-1) + r) \cdot {{^{20}{C_r}}}$ We can split this sum into two parts: $S = \sum_{r=0}^{20} r(r-1) \cdot {{^{20}{C_r}}} + \sum_{r=0}^{20} r \cdot {{^{20}{C_r}}}$

**Step 2: Evaluate the second sum, $\sum_{r=0}^{20} r \cdot {{^{20}{C_r}}}$} For this sum, we use the identity $r \cdot {{n}{C_r}} = n \cdot {{n-1}{C_{r-1}}}$. Here $n=20$. So, $r \cdot {{^{20}{C_r}}} = 20 \cdot {{^{19}{C_{r-1}}}}$. Note that for $r=0$, the term $r \cdot {{^{20}{C_r}}} = 0 \cdot {{^{20}{C_0}}} = 0$. So the sum effectively starts from $r=1$. $\sum_{r=0}^{20} r \cdot {{^{20}{C_r}}} = \sum_{r=1}^{20} 20 \cdot {{^{19}{C_{r-1}}}}$ Let $k = r-1$. When $r=1, k=0$. When $r=20, k=19$. $= 20 \sum_{k=0}^{19} {{^{19}{C_k}}}$ Using the sum of binomial coefficients formula $\sum_{k=0}^n {{n}{C_k}} = 2^n$: $= 20 \cdot 2^{19}$

**Step 3: Evaluate the first sum, $\sum_{r=0}^{20} r(r-1) \cdot {{^{20}{C_r}}}$} For this sum, we use the identity $r(r-1) \cdot {{n}{C_r}} = n(n-1) \cdot {{n-2}{C_{r-2}}}$. Here $n=20$. So, $r(r-1) \cdot {{^{20}{C_r}}} = 20(20-1) \cdot {{^{20-2}{C_{r-2}}}} = 20 \cdot 19 \cdot {{^{18}{C_{r-2}}}}$. Note that for $r=0$ and $r=1$, the term $r(r-1) \cdot {{^{20}{C_r}}}$ is $0$. So the sum effectively starts from $r=2$. $\sum_{r=0}^{20} r(r-1) \cdot {{^{20}{C_r}}} = \sum_{r=2}^{20} 20 \cdot 19 \cdot {{^{18}{C_{r-2}}}}$ Let $k = r-2$. When $r=2, k=0$. When $r=20, k=18$. $= 20 \cdot 19 \sum_{k=0}^{18} {{^{18}{C_k}}}$ Using the sum of binomial coefficients formula: $= 20 \cdot 19 \cdot 2^{18}$ $= 380 \cdot 2^{18}$

Step 4: Combine the results Now, substitute the results from Step 2 and Step 3 back into the expression for $S$: $S = (380 \cdot 2^{18}) + (20 \cdot 2^{19})$ To combine these terms, it's helpful to express them with the same power of 2. We know that $2^{19} = 2 \cdot 2^{18}$. $S = 380 \cdot 2^{18} + 20 \cdot (2 \cdot 2^{18})$ $S = 380 \cdot 2^{18} + 40 \cdot 2^{18}$ Now, factor out $2^{18}$: $S = (380 + 40) \cdot 2^{18}$ $S = 420 \cdot 2^{18}$

Alternative Method: Using Differentiation

Another powerful technique to evaluate such sums is by differentiating the binomial expansion. We know that: $(1+x)^n = \sum_{r=0}^n {{n}{C_r}} x^r$ Step A: Differentiate once with respect to $x$ $n(1+x)^{n-1} = \sum_{r=0}^n r {{n}{C_r}} x^{r-1}$ Step B: Multiply by $x$ $n x (1+x)^{n-1} = \sum_{r=0}^n r {{n}{C_r}} x^r$ Step C: Differentiate again with respect to $x$ The left side (LHS) is $n \cdot \frac{d}{dx} [x (1+x)^{n-1}]$. Using the product rule: $LHS = n [(1+x)^{n-1} \cdot 1 + x \cdot (n-1)(1+x)^{n-2}]$ $LHS = n(1+x)^{n-1} + n(n-1)x(1+x)^{n-2}$ The right side (RHS) is $\sum_{r=0}^n r^2 {{n}{C_r}} x^{r-1}$. So, we have: $\sum_{r=0}^n r^2 {{n}{C_r}} x^{r-1} = n(1+x)^{n-1} + n(n-1)x(1+x)^{n-2}$ Step D: Set $x=1$ Setting $x=1$ allows us to remove the $x$ terms and get the desired sum. $\sum_{r=0}^n r^2 {{n}{C_r}} = n(1+1)^{n-1} + n(n-1)(1)(1+1)^{n-2}$ $\sum_{r=0}^n r^2 {{n}{C_r}} = n \cdot 2^{n-1} + n(n-1) \cdot 2^{n-2}$ Now, substitute $n=20$: $\sum_{r=0}^{20} r^2 {{^{20}{C_r}}} = 20 \cdot 2^{20-1} + 20(20-1) \cdot 2^{20-2}$ $= 20 \cdot 2^{19} + 20 \cdot 19 \cdot 2^{18}$ $= 20 \cdot (2 \cdot 2^{18}) + 380 \cdot 2^{18}$ $= 40 \cdot 2^{18} + 380 \cdot 2^{18}$ $= (40 + 380) \cdot 2^{18}$ $= 420 \cdot 2^{18}$

Both methods yield the same result, confirming its correctness.

Comparing with Options The calculated value is $420 \times 2^{18}$. Let's check the given options: (A) $420 \times {2^{19}} = 420 \times 2 \times 2^{18} = 840 \times 2^{18}$ (B) $380 \times {2^{19}} = 380 \times 2 \times 2^{18} = 760 \times 2^{18}$ (C) $380 \times {2^{18}}$ (D) $420 \times {2^{18}}$

Our result $420 \times 2^{18}$ matches option (D).

Important Tips and Common Mistakes

• Understanding Identities: Memorize or be able to quickly derive the identities for $r \cdot {{n}{C_r}}$ and $r(r-1) \cdot {{n}{C_r}}$. They are fundamental for simplifying such sums.
• Lower Limits of Summation: Be careful when applying these identities. Terms like $r=0$ or $r=1$ often result in zero for $r \cdot {{n}{C_r}}$ or $r(r-1) \cdot {{n}{C_r}}$, which means the effective starting point of the sum changes.
• Algebraic Simplification: Pay attention to simplifying powers of 2 (e.g., $2^{19} = 2 \cdot 2^{18}$) to combine terms effectively and match the format of the options.
• Differentiation Method: The differentiation method is a powerful general approach for sums involving $r$, $r^2$, etc. in binomial expansions. Practice this method to ensure accuracy. It serves as a great way to verify results obtained through other methods.

Summary/Key Takeaway

This problem demonstrates the effective use of binomial coefficient identities to simplify complex sums. By decomposing $r^2$ into $r(r-1) + r$, we transformed the original sum into two simpler sums that could be directly evaluated using the identities $r \cdot {{n}{C_r}} = n \cdot {{n-1}{C_{r-1}}}$ and $r(r-1) \cdot {{n}{C_r}} = n(n-1) \cdot {{n-2}{C_{r-2}}}$, along with the sum of binomial coefficients $2^n$. The differentiation method provides an elegant alternative and a good cross-check for the result. The final answer is $420 \times 2^{18}$.