1. Key Concepts and Formulas

This problem leverages fundamental identities related to binomial coefficients. Understanding these properties is crucial for simplifying complex summations efficiently.

• Binomial Theorem Expansion: The expansion of $(1+x)^n$ is given by: $(1+x)^n = \sum_{k=0}^n {^n{C_k}} x^k = {^n{C_0}} + {^n{C_1}}x + {^n{C_2}}x^2 + \dots + {^n{C_n}}x^n$

• Sum of Binomial Coefficients: By setting $x=1$ in the binomial expansion, we obtain the sum of all binomial coefficients: $\sum_{k=0}^n {^n{C_k}} = {2^n}$ Explanation: This identity signifies that the sum of all coefficients in the expansion of $(1+x)^n$ is $2^n$.

• Property $k \cdot {^n{C_k}}$: This identity is extremely useful when a summation term involves the product of the index $k$ and a binomial coefficient. It can be derived from the definition of ${^n{C_k}} = \frac{n!}{k!(n-k)!}$: $k \cdot {^n{C_k}} = k \cdot \frac{n!}{k!(n-k)!} = \frac{n!}{(k-1)!(n-k)!}$ We can rewrite the right side to match the form of a binomial coefficient: $= n \cdot \frac{(n-1)!}{(k-1)!((n-1)-(k-1))!} = n \cdot {^{n-1}{C_{k-1}}}$ So, for $k \ge 1$: $k \cdot {^n{C_k}} = n \cdot {^{n-1}{C_{k-1}}}$ Explanation: This property allows us to "reduce" the 'n' and 'k' in the binomial coefficient and extract 'n' as a factor. Note that for $k=0$, the term $k \cdot {^n{C_k}}$ is $0 \cdot {^n{C_0}} = 0$. This implies that when a summation starts from $k=0$ and involves $k \cdot {^n{C_k}}$, we can effectively start the summation from $k=1$ without changing the total sum, making the application of this property straightforward.

2. Decomposing the Given Summation

The problem requires us to evaluate the following summation: $S = \sum\limits_{k = 0}^{10} {({2^2} + 3k){}^{10}{C_k}}$

Step 1: Simplify the constant term and split the summation. We begin by simplifying ${2^2}$ to $4$. Then, we can use the linearity property of summation to split the expression into two separate sums: $S = \sum\limits_{k = 0}^{10} {(4 + 3k){}^{10}{C_k}}$ $S = \sum\limits_{k = 0}^{10} {4 \cdot {}^{10}{C_k}} + \sum\limits_{k = 0}^{10} {3k \cdot {}^{10}{C_k}}$

Step 2: Factor out constants. Constants can be moved outside the summation sign: $S = 4\sum\limits_{k = 0}^{10} {{}^{10}{C_k}} + 3\sum\limits_{k = 0}^{10} {k \cdot {}^{10}{C_k}}$ Now we have two distinct summations to evaluate.

3. Evaluating the First Summation

Consider the first part: $4\sum\limits_{k = 0}^{10} {{}^{10}{C_k}}$. Step 3: Apply the Sum of Binomial Coefficients identity. Using the formula $\sum_{k=0}^n {^n{C_k}} = 2^n$ with $n=10$: $\sum\limits_{k = 0}^{10} {{}^{10}{C_k}} = {2^{10}}$ So, the first part of our expression becomes: $4 \cdot {2^{10}}$

4. Evaluating the Second Summation

Now let's evaluate the second part: $3\sum\limits_{k = 0}^{10} {k \cdot {}^{10}{C_k}}$. Step 4: Apply the property $k \cdot {^n{C_k}} = n \cdot {^{n-1}{C_{k-1}}}$. As discussed in the Key Concepts, the term for $k=0$ ($0 \cdot {}^{10}{C_0}$) is zero, so we can change the lower limit of summation from $k=0$ to $k=1$: $3\sum\limits_{k = 0}^{10} {k \cdot {}^{10}{C_k}} = 3\sum\limits_{k = 1}^{10} {k \cdot {}^{10}{C_k}}$ Now, apply the property $k \cdot {^n{C_k}} = n \cdot {^{n-1}{C_{k-1}}}$ with $n=10$: $3\sum\limits_{k = 1}^{10} {10 \cdot {}^{10-1}{C_{k-1}}}$ $= 3\sum\limits_{k = 1}^{10} {10 \cdot {}^{9}{C_{k-1}}}$ Factor out the constant $10$: $= 3 \cdot 10 \sum\limits_{k = 1}^{10} {{}^{9}{C_{k-1}}}$ $= 30 \sum\limits_{k = 1}^{10} {{}^{9}{C_{k-1}}}$

Step 5: Change the index of summation to apply the Sum of Binomial Coefficients identity. Let $j = k-1$. When $k=1$, $j=0$. When $k=10$, $j=9$. The summation transforms into: $30 \sum\limits_{j = 0}^{9} {{}^{9}{C_j}}$ Now, apply the Sum of Binomial Coefficients identity $\sum_{j=0}^n {^n{C_j}} = 2^n$ with $n=9$: $\sum\limits_{j = 0}^{9} {{}^{9}{C_j}} = {2^9}$ So, the second part of our expression evaluates to: $30 \cdot {2^9}$

Tip: Always be careful when changing the index of summation. Ensure the new limits and the general term correctly reflect the original summation. A common mistake is to forget to adjust the limits.

5. Combining the Evaluated Sums

Now we substitute the results from steps 3 and 5 back into our decomposed summation for $S$: $S = 4 \cdot {2^{10}} + 30 \cdot {2^9}$

Step 6: Simplify the expression by unifying powers of 2. To combine these terms, we express them with a common power of 2. We know that ${2^{10}} = 2 \cdot {2^9}$, or conversely, ${2^9} = \frac{1}{2} \cdot {2^{10}}$. Let's convert $30 \cdot {2^9}$ to a term involving ${2^{10}}$: $30 \cdot {2^9} = 15 \cdot (2 \cdot {2^9}) = 15 \cdot {2^{10}}$ Substitute this back into the expression for $S$: $S = 4 \cdot {2^{10}} + 15 \cdot {2^{10}}$ Now, factor out the common term ${2^{10}}$: $S = (4 + 15) \cdot {2^{10}}$ $S = 19 \cdot {2^{10}}$

6. Determining $\alpha$ and $\beta$

The problem states that the given sum is equal to $\alpha \cdot {3^{10}} + \beta \cdot {2^{10}}$. We have found that the sum evaluates to $19 \cdot {2^{10}}$. Therefore, we set up the equality: $19 \cdot {2^{10}} = \alpha \cdot {3^{10}} + \beta \cdot {2^{10}}$

Step 7: Compare coefficients. For this equation to hold true for specific numerical bases ${3^{10}}$ and ${2^{10}}$ (which are distinct values), the coefficients of each base must match on both sides. Since there is no ${3^{10}}$ term on the left side, its coefficient must be zero: $\alpha = 0$ Then, comparing the coefficients of ${2^{10}}$: $19 = \beta$ Thus, we have found: $\alpha = 0 \quad \text{and} \quad \beta = 19$

7. Calculating $\alpha + \beta$

Finally, we calculate the required sum: $\alpha + \beta = 0 + 19 = 19$

8. Summary and Key Takeaway

This problem is an excellent illustration of how algebraic manipulation combined with fundamental binomial coefficient identities can simplify complex summations. The key steps involved:

1. Decomposing the original sum into simpler parts.
2. Applying the identity $\sum_{k=0}^n {^n{C_k}} = 2^n$.
3. Utilizing the property $k \cdot {^n{C_k}} = n \cdot {^{n-1}{C_{k-1}}}$ along with a careful change of summation index.
4. Consolidating terms by ensuring a common base for powers.
5. Equating the final simplified expression to the given form to determine unknown coefficients.

Mastering these techniques is essential for solving a wide range of problems in combinatorics and the binomial theorem.

Note: Based on the step-by-step derivation using standard binomial identities, the calculated value for $\alpha + \beta$ is $19$. The "Correct Answer: 0" provided in the problem statement appears to be inconsistent with the problem setup and standard mathematical results.