Elaborate Solution for Binomial Summation

Problem Statement: Given the sum $S = {1^2}\,{}^{20}{C_1} + {2^2}\,{}^{20}{C_2} + {3^2}\,{}^{20}{C_3} + \ldots + {20^2}\,{}^{20}{C_{20}}$. If $S = A(2^\beta)$, determine the ordered pair $(A, \beta)$.

Key Concepts and Formulas: To solve this problem, we will utilize fundamental identities involving binomial coefficients:

1. Identity 1: $r \binom{n}{r} = n \binom{n-1}{r-1}$
   • Explanation: This identity allows us to reduce both the factor of $r$ and the upper index of the binomial coefficient. It is derived from the definition of $\binom{n}{r} = \frac{n!}{r!(n-r)!}$.
   • $r \binom{n}{r} = r \cdot \frac{n!}{r!(n-r)!} = \frac{n!}{(r-1)!(n-r)!} = n \cdot \frac{(n-1)!}{(r-1)!((n-1)-(r-1))!} = n \binom{n-1}{r-1}$.
2. Identity 2: $\sum_{k=0}^{n} \binom{n}{k} = 2^n$
   • Explanation: This is the sum of all binomial coefficients for a given $n$, which arises from the binomial expansion of $(1+1)^n$.

Step-by-Step Solution:

Step 1: Express the sum in summation notation. The given sum can be written concisely using summation notation: $S = \sum_{r=1}^{20} r^2 \binom{20}{r}$

• Explanation: The sum starts from $r=1$ because the term corresponding to $r=0$ would be $0^2 \binom{20}{0} = 0$, which does not contribute to the sum. This notation simplifies subsequent manipulations.

Step 2: Apply Identity 1 to simplify $r^2 \binom{20}{r}$. We can rewrite $r^2 \binom{20}{r}$ as $r \cdot \left(r \binom{20}{r}\right)$. Applying Identity 1 with $n=20$: $r \binom{20}{r} = 20 \binom{19}{r-1}$ Substituting this back into the sum: $S = \sum_{r=1}^{20} r \left(20 \binom{19}{r-1}\right) = 20 \sum_{r=1}^{20} r \binom{19}{r-1}$

• Explanation: This step is crucial as it reduces the complexity by lowering the upper index of the binomial coefficient from 20 to 19. It effectively deals with one power of $r$.

Step 3: Manipulate the term $r \binom{19}{r-1}$ to apply Identity 1 again. To apply Identity 1 to $r \binom{19}{r-1}$, we need the term multiplying the binomial coefficient to match its lower index. Here, the lower index is $r-1$. We achieve this by rewriting $r$ as $(r-1) + 1$: $S = 20 \sum_{r=1}^{20} ((r-1) + 1) \binom{19}{r-1}$ Now, we can split the sum into two parts: $S = 20 \left[ \sum_{r=1}^{20} (r-1) \binom{19}{r-1} + \sum_{r=1}^{20} \binom{19}{r-1} \right]$

• Explanation: This strategic separation allows us to apply Identity 1 to the first sum (where the factor is $r-1$, matching the lower index) and directly evaluate the second sum using Identity 2.

Step 4: Evaluate each of the two sums separately.

• First Sum: $\sum_{r=1}^{20} (r-1) \binom{19}{r-1}$

  • For $r=1$, the term $(r-1)$ is $0$, so the term is $0 \cdot \binom{19}{0} = 0$. We can safely start the sum from $r=2$.
  • Let $k = r-1$. When $r=2$, $k=1$. When $r=20$, $k=19$.
  • The sum becomes $\sum_{k=1}^{19} k \binom{19}{k}$.
  • Apply Identity 1 with $n=19$: $k \binom{19}{k} = 19 \binom{18}{k-1}$.
  • So, the sum is $\sum_{k=1}^{19} 19 \binom{18}{k-1} = 19 \sum_{k=1}^{19} \binom{18}{k-1}$.
  • Let $j = k-1$. When $k=1$, $j=0$. When $k=19$, $j=18$.
  • The sum becomes $19 \sum_{j=0}^{18} \binom{18}{j}$.
  • Using Identity 2, $\sum_{j=0}^{18} \binom{18}{j} = 2^{18}$.
  • Thus, the First Sum evaluates to $19 \cdot 2^{18}$.
  • Explanation: We performed a change of variable ($k=r-1$, then $j=k-1$) to correctly apply the identities and align summation limits. Each step reduces the binomial coefficient's upper index and simplifies the factor multiplying it.

• Second Sum: $\sum_{r=1}^{20} \binom{19}{r-1}$

  • Let $k = r-1$. When $r=1$, $k=0$. When $r=20$, $k=19$.
  • The sum becomes $\sum_{k=0}^{19} \binom{19}{k}$.
  • Using Identity 2, $\sum_{k=0}^{19} \binom{19}{k} = 2^{19}$.
  • Thus, the Second Sum evaluates to $2^{19}$.
  • Explanation: This is a direct application of the sum of binomial coefficients property after a simple change of variable to make the summation indices clear.

Step 5: Substitute the evaluated sums back into the expression for $S$. Substitute the results from Step 4 into the expression from Step 3: $S = 20 \left[ (19 \cdot 2^{18}) + 2^{19} \right]$

• Explanation: This brings together the simplified parts of the original sum.

Step 6: Simplify the expression to the form $A(2^\beta)$. To express $S$ in the desired form, we factor out common terms. Notice that $2^{19}$ can be written as $2 \cdot 2^{18}$: $S = 20 \left[ 19 \cdot 2^{18} + 2 \cdot 2^{18} \right]$ Factor out $2^{18}$ from the terms inside the bracket: $S = 20 \cdot 2^{18} (19 + 2)$ $S = 20 \cdot 2^{18} (21)$ Rearranging the terms: $S = (20 \cdot 21) \cdot 2^{18}$ $S = 420 \cdot 2^{18}$

• Explanation: This is purely algebraic simplification. Factoring out $2^{18}$ is a key step to consolidate the terms and arrive at the $A(2^\beta)$ format.

Step 7: Identify A and $\beta$. Comparing our result $S = 420 \cdot 2^{18}$ with the given form $S = A \cdot 2^\beta$, we can identify the values: $A = 420$ $\beta = 18$ Therefore, the ordered pair $(A, \beta)$ is $(420, 18)$.

Tips for Success & Common Pitfalls:

• Master Binomial Identities: Memorize and understand the application of key identities like $r \binom{n}{r} = n \binom{n-1}{r-1}$ and $\sum \binom{n}{r} = 2^n$. These are foundational for solving problems involving sums of binomial coefficients.
• Careful with Summation Limits: When performing a change of variable (e.g., $k=r-1$) or when terms become zero (e.g., $k \binom{n}{k} = 0$ for $k=0$), always adjust the summation limits precisely.
• Strategic Manipulation: Problems often require breaking down terms (e.g., $r^2$ into $r \cdot r$ or $r$ into $(r-1)+1$) to enable further application of identities. Think ahead about how each transformation will allow for subsequent simplifications.
• Factoring: Be proficient in factoring common terms, especially powers of 2, to simplify the final expression and match the required format.

Summary & Key Takeaway: This problem demonstrates a systematic approach to evaluating complex sums involving $r^k \binom{n}{r}$ by repeatedly applying the identity $r \binom{n}{r} = n \binom{n-1}{r-1}$. The strategy involves gradually reducing the power of $r$ and the upper index of the binomial coefficient until the expression can be simplified using the fundamental sum $\sum \binom{m}{j} = 2^m$. This method is a powerful tool for solving various problems in binomial theorem.