If $\sum_{r=0}^{10} \left(\frac{10^{r+1} - 1}{10^r}\right) \cdot {}^{11}C_{r+1} = \frac{\alpha^{11} - 11^{11}}{10^{10}}$, then $\alpha$ is equal to:

Step 1: Simplify the general term of the summation

First, we simplify the expression inside the summation. We can split the fraction $\frac{10^{r+1} - 1}{10^r}$ into two terms: $\frac{10^{r+1}}{10^r} - \frac{1}{10^r}$. This simplifies to $10 - \frac{1}{10^r}$.

Step 2: Substitute the simplified term into the summation

Now we substitute the simplified term back into the summation. We can then split the summation into two separate summations due to the distributive property.

Step 3: Adjust the index of summation for the first term

For the first summation, let $k = r+1$. When $r=0$, $k=1$. When $r=10$, $k=11$. So the summation becomes $10 \sum_{k=1}^{11} {}^{11}C_k$. We know that $\sum_{k=0}^{n} {}^{n}C_k = 2^n$, so $\sum_{k=1}^{n} {}^{n}C_k = 2^n - {}^{n}C_0 = 2^n - 1$. Therefore, $10 \sum_{k=1}^{11} {}^{11}C_k = 10(2^{11} - 1)$.

Step 4: Adjust the index of summation for the second term

For the second summation, again let $k = r+1$. The summation becomes $\sum_{k=1}^{11} \frac{1}{10^{k-1}} {}^{11}C_k$. We can rewrite $\frac{1}{10^{k-1}}$ as $10 \cdot \frac{1}{10^k}$. So the term is $10 \sum_{k=1}^{11} {}^{11}C_k \left(\frac{1}{10}\right)^k$.

Step 5: Apply the binomial theorem to the second term

Recall the binomial theorem: $(a+b)^n = \sum_{k=0}^{n} {}^{n}C_k a^{n-k} b^k$. Here, we have $a=1$, $b=\frac{1}{10}$, and $n=11$. So, $\sum_{k=0}^{11} {}^{11}C_k \left(\frac{1}{10}\right)^k = \left(1 + \frac{1}{10}\right)^{11}$. Since our summation starts from $k=1$, we subtract the $k=0$ term, which is ${}^{11}C_0 \left(\frac{1}{10}\right)^0 = 1 \cdot 1 = 1$. Thus, the second term simplifies to $10 \left[ \left(\frac{11}{10}\right)^{11} - 1 \right]$.

Step 6: Combine the two terms and solve for $\alpha$

Now we combine the simplified first and second terms. $10(2^{11} - 1) - 10 \left[ \left(\frac{11}{10}\right)^{11} - 1 \right] = 10 \cdot 2^{11} - 10 - 10 \cdot \frac{11^{11}}{10^{11}} + 10$. The $-10$ and $+10$ cancel out. The term $10 \cdot \frac{11^{11}}{10^{11}}$ simplifies to $\frac{11^{11}}{10^{10}}$. So the entire expression is $10 \cdot 2^{11} - \frac{11^{11}}{10^{10}}$. We are given that this is equal to $\frac{\alpha^{11} - 11^{11}}{10^{10}}$. We can rewrite $10 \cdot 2^{11}$ as $\frac{10^{11} \cdot 2^{11}}{10^{10}} = \frac{(10 \cdot 2)^{11}}{10^{10}} = \frac{20^{11}}{10^{10}}$. Therefore, $\frac{20^{11} - 11^{11}}{10^{10}} = \frac{\alpha^{11} - 11^{11}}{10^{10}}$. Comparing the numerators, we get $\alpha^{11} = 20^{11}$, which implies $\alpha = 20$.

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