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 :

1. Simplify the General Term

We begin by simplifying the fraction inside the summation:

$$\frac{10^{r+1}-1}{10^r} = \frac{10^{r+1}}{10^r} - \frac{1}{10^r} = 10 - 10^{-r}$$

Substituting this back into the sum, we can split it into two separate series:

$$S = \sum_{r=0}^{10} \left(10 - 10^{-r}\right) {}^{11}C_{r+1} = 10\sum_{r=0}^{10} {}^{11}C_{r+1} - \sum_{r=0}^{10} 10^{-r} \cdot {}^{11}C_{r+1}$$

To make things easier, let's use a change of variable. Let $$k = r + 1$$. As $$r$$ varies from $$0$$ to $$10$$, $$k$$ runs from $$1$$ to $$11$$:

$$
\begin{aligned}
S = 10\sum_{k=1}^{11} {}^{11}C_k - \sum_{k=1}^{11} 10^{-(k-1)} {}^{11}C_k && \text{(1)}
\end{aligned}
$$

2. Evaluate the First Summation

Using the standard binomial theorem identity $$\sum_{k=0}^{11} {}^{11}C_k = 2^{11}$$, we subtract the missing $$k=0$$ term ($${}^{11}C_0 = 1$$):

$$\sum_{k=1}^{11} {}^{11}C_k = 2^{11} - {}^{11}C_0 = 2^{11} - 1$$

Multiplying by the constant factor of $$10$$:

$$10\sum_{k=1}^{11} {}^{11}C_k = 10\left(2^{11} - 1\right) = 10(2048 - 1) = 20470 \tag{2}$$

3. Evaluate the Second Summation

We rewrite the exponent $$10^{-(k-1)}$$ as $$10 \cdot 10^{-k}$$:

$$\sum_{k=1}^{11} 10^{-(k-1)} {}^{11}C_k = 10\sum_{k=1}^{11} 10^{-k} {}^{11}C_k$$

Using the binomial expansion formula $$(1+x)^{11} = \sum_{k=0}^{11} {}^{11}C_k x^k$$ with $$x = \frac{1}{10}$$:

$$\sum_{k=1}^{11} {}^{11}C_k 10^{-k} = \left(1 + \frac{1}{10}\right)^{11} - {}^{11}C_0 = \left(\frac{11}{10}\right)^{11} - 1$$

Now, multiply this by $$10$$:

$$\sum_{k=1}^{11} 10^{-(k-1)} {}^{11}C_k = 10\left[\left(\frac{11}{10}\right)^{11} - 1\right] = 10 \cdot \frac{11^{11}}{10^{11}} - 10 = \frac{11^{11}}{10^{10}} - 10 \tag{3}$$

4. Combine and Solve for $$\alpha$$

Substitute equation (2) and equation (3) back into the primary expression for $$S$$ (Equation 1):

$$S = 20470 - \left( \frac{11^{11}}{10^{10}} - 10 \right)$$ $$S = 20470 + 10 - \frac{11^{11}}{10^{10}}$$  $$S = 20480 - \frac{11^{11}}{10^{10}}$$

To match the target format given in the problem, let's establish a common denominator of $$10^{10}$$:

$$S = \frac{20480 \cdot 10^{10} - 11^{11}}{10^{10}}$$

We are given that:

$$S = \frac{\alpha^{11} - 11^{11}}{10^{10}}$$

By comparing the numerators directly:

$$\alpha^{11} = 20480 \cdot 10^{10}$$

Since $$2048 = 2^{11}$$, we can rewrite $$20480$$ as $$2^{11} \cdot 10$$:

$$\alpha^{11} = (2^{11} \cdot 10) \cdot 10^{10}$$  $$\alpha^{11} = 2^{11} \cdot 10^{11}$$  $$\alpha^{11} = (2 \cdot 10)^{11} = 20^{11}$$

Taking the $$11\text{-th}$$ root on both sides:

$$\alpha = 20$$

Final Answer

Option D (20)