Key Concept: Binomial Theorem and its Applications

This problem primarily utilizes the Binomial Theorem, which states that for any non-negative integer $n$, $(x+y)^n = \sum_{k=0}^{n} {}^{n}C_k x^{n-k} y^k$ A common variant used here is for $(1+x)^n = \sum_{k=0}^{n} {}^{n}C_k x^k$. Additionally, we use the property that the sum of binomial coefficients $\sum_{k=0}^{n} {}^{n}C_k = 2^n$.

Problem Statement We need to find the value of $\alpha$ given the equation: \sum_{r = 0}^{10} { \left({{{{10}^{r + 1}} - 1} \over {{{10}^r}}}).{}^{11}{C_{r + 1}} = {{{}_{\alpha} 11 - {{11}^{11}}} \over {{{10}^{10}}}} } The notation ${}_\alpha 11$ is ambiguous. Based on the options and the standard way such problems are posed, it is interpreted as $\alpha^{11}$. So, the target form is $\frac{\alpha^{11} - 11^{11}}{10^{10}}$.

Step-by-Step Solution

1. Simplify the general term of the summation. The term inside the summation is $\left(\frac{10^{r+1} - 1}{10^r}\right) \cdot {}^{11}{C_{r+1}}$. Let's simplify the fractional part: $\frac{10^{r+1} - 1}{10^r} = \frac{10^{r+1}}{10^r} - \frac{1}{10^r} = 10 - \frac{1}{10^r}$ Explanation: We separate the fraction into two terms to make it easier to work with. This is a standard algebraic simplification.

2. Rewrite the $\frac{1}{10^r}$ term for binomial expansion compatibility. Observe that $\frac{1}{10^r}$ can be expressed in terms of $\left(\frac{1}{10}\right)^{r+1}$: $\frac{1}{10^r} = 10 \cdot \frac{1}{10^{r+1}} = 10 \left(\frac{1}{10}\right)^{r+1}$ Explanation: This step is crucial for aligning the exponent of $\left(\frac{1}{10}\right)$ with the index of the binomial coefficient ${}^{11}{C_{r+1}}$ (where the index is $r+1$). This transformation allows us to use the binomial theorem directly in the next steps.

3. Substitute the simplified term back into the summation and separate. Now, the original summation becomes: $\sum_{r=0}^{10} \left(10 - 10\left(\frac{1}{10}\right)^{r+1}\right) {}^{11}{C_{r+1}}$ $= 10 \sum_{r=0}^{10} {}^{11}{C_{r+1}} - 10 \sum_{r=0}^{10} {}^{11}{C_{r+1}} \left(\frac{1}{10}\right)^{r+1}$ Explanation: We distribute the $10$ and split the summation into two parts, as linearity of summation allows this. This makes each part independently solvable using known binomial identities.

4. Evaluate the first summation. Let $S_1 = 10 \sum_{r=0}^{10} {}^{11}{C_{r+1}}$. Let $k = r+1$. When $r=0$, $k=1$. When $r=10$, $k=11$. So the sum becomes: $S_1 = 10 \sum_{k=1}^{11} {}^{11}{C_k} = 10 \left( {}^{11}{C_1} + {}^{11}{C_2} + \ldots + {}^{11}{C_{11}} \right)$ We know from the Binomial Theorem that $\sum_{k=0}^{11} {}^{11}{C_k} = {}^{11}{C_0} + {}^{11}{C_1} + \ldots + {}^{11}{C_{11}} = (1+1)^{11} = 2^{11}$. Therefore, ${}^{11}{C_1} + {}^{11}{C_2} + \ldots + {}^{11}{C_{11}} = 2^{11} - {}^{11}{C_0} = 2^{11} - 1$. Substituting this back: $S_1 = 10 (2^{11} - 1) = 10 \cdot 2^{11} - 10$ Explanation: We perform a change of index to align with the standard form of binomial sums. Then, we use the identity $\sum_{k=0}^n {}^n C_k = 2^n$ to evaluate the sum of binomial coefficients, adjusting for the missing ${}^{11}C_0$ term.

5. Evaluate the second summation. Let $S_2 = 10 \sum_{r=0}^{10} {}^{11}{C_{r+1}} \left(\frac{1}{10}\right)^{r+1}$. Again, let $k = r+1$. The sum becomes: $S_2 = 10 \sum_{k=1}^{11} {}^{11}{C_k} \left(\frac{1}{10}\right)^k$ Let $x = \frac{1}{10}$. The sum inside the brackets is: ${}^{11}{C_1}x^1 + {}^{11}{C_2}x^2 + \ldots + {}^{11}{C_{11}}x^{11}$ From the Binomial Theorem, $(1+x)^{11} = {}^{11}{C_0}x^0 + {}^{11}{C_1}x^1 + \ldots + {}^{11}{C_{11}}x^{11}$. So, $\sum_{k=1}^{11} {}^{11}{C_k}x^k = (1+x)^{11} - {}^{11}{C_0}x^0 = (1+x)^{11} - 1$. Substitute $x = \frac{1}{10}$: $S_2 = 10 \left[ \left(1 + \frac{1}{10}\right)^{11} - 1 \right] = 10 \left[ \left(\frac{11}{10}\right)^{11} - 1 \right]$ $S_2 = 10 \left( \frac{11^{11}}{10^{11}} - 1 \right) = \frac{10 \cdot 11^{11}}{10^{11}} - 10 = \frac{11^{11}}{10^{10}} - 10$ Explanation: Similar to the first sum, we adjust the index. We then recognize the pattern as part of the binomial expansion of $(1+x)^{11}$. By subtracting the missing ${}^{11}C_0$ term, we can express the sum using the binomial expansion.

6. Combine the results of the two summations. The total sum is $S = S_1 - S_2$: $S = (10 \cdot 2^{11} - 10) - \left( \frac{11^{11}}{10^{10}} - 10 \right)$ $S = 10 \cdot 2^{11} - 10 - \frac{11^{11}}{10^{10}} + 10$ $S = 10 \cdot 2^{11} - \frac{11^{11}}{10^{10}}$ Explanation: We substitute the calculated values of $S_1$ and $S_2$ and simplify the expression.

7. Express the result in the target format. The target format is $\frac{\alpha^{11} - 11^{11}}{10^{10}}$. We need to rewrite $10 \cdot 2^{11}$ to match the numerator's structure. $10 \cdot 2^{11} = \frac{10^{10} \cdot (10 \cdot 2^{11})}{10^{10}} = \frac{10^{11} \cdot 2^{11}}{10^{10}}$ This can be written as: $\frac{(10 \cdot 2)^{11}}{10^{10}} = \frac{20^{11}}{10^{10}}$ So, the total sum is: $S = \frac{20^{11}}{10^{10}} - \frac{11^{11}}{10^{10}} = \frac{20^{11} - 11^{11}}{10^{10}}$ Explanation: We manipulate the term $10 \cdot 2^{11}$ to fit the structure of the numerator, specifically to form a base raised to the power of 11. This reveals the value of $\alpha$.

8. Determine the value of $\alpha$. Comparing our result $\frac{20^{11} - 11^{11}}{10^{10}}$ with the given form $\frac{\alpha^{11} - 11^{11}}{10^{10}}$, we find that: $\alpha = 20$

Common Mistakes & Tips

• Index Alignment: Be extremely careful with the starting and ending indices of summations and binomial coefficients. A common mistake is to overlook a ${}^{n}C_0$ term or an offset in the index.
• Algebraic Simplification: Ensure that the initial algebraic simplification of the general term is correct. Errors here propagate through the entire solution.
• Binomial Theorem Application: Remember the exact form of the binomial theorem. If terms are missing (e.g., ${}^{n}C_0$), adjust the expansion accordingly.
• Handling Powers: Pay close attention to how powers of 10 are combined and simplified, especially when dealing with denominators. $(a \cdot b)^n = a^n \cdot b^n$.

Summary By simplifying the general term, strategically rewriting it to align with binomial expansion forms, and carefully evaluating the resulting summations, we found the value of the expression. Comparing this to the given structure, we determined $\alpha = 20$. The key steps involved recognizing and applying the Binomial Theorem, alongside meticulous algebraic manipulation.

The final answer is $\boxed{\text{20}}$. However, since the provided correct answer is A (11), there seems to be a discrepancy between the problem statement (or its intended interpretation) and the options/correct answer. Based on the given problem statement and standard mathematical rules, the derived answer is 20. If 11 is indeed the correct answer, the problem statement would likely be subtly different, possibly involving a negative sign or a different power for $\alpha$. Without further clarification, following the steps from the given problem leads to $\alpha=20$.