If $$(10)^9 + 2(11)^1(10)^8 + 3(11)^2(10)^7 + \ldots + 10(11)^9 = k(10)^9$$, then $$k$$ is equal to:

We are asked to evaluate the expression

$$ (10)^9 + 2(11)^1(10)^8 + 3(11)^2(10)^7 + \ldots + 10(11)^9 = k(10)^9 $$

and then find the constant $$k$$. Observe carefully that every term contains a power of $$10$$ whose exponent decreases by 1 while the coefficient in front of the power of $$11$$ increases by 1. To write this compactly, let us introduce an index $$r$$ starting from 0:

$$ \text{Term for }r=0:\;(r+1)(11)^r(10)^{9-r} = 1\cdot 11^0\cdot 10^9 $$ $$ \text{Term for }r=1:\;(r+1)(11)^r(10)^{9-r} = 2\cdot 11^1\cdot 10^8 $$ $$ \;\;\vdots $$ $$ \text{Term for }r=9:\;(r+1)(11)^r(10)^{9-r} = 10\cdot 11^9\cdot 10^0 $$

Hence the whole sum can be written as

$$ S \;=\; \sum_{r=0}^{9} (r+1)\,11^r\,10^{\,9-r}. $$

We notice that every term possesses a common factor $$10^9$$ because $$10^{\,9-r} = 10^9\,(10^{-1})^{r}$$. Taking this common factor outside gives

$$ S \;=\; 10^9 \sum_{r=0}^{9} (r+1)\Bigl(\frac{11}{10}\Bigr)^{\!r}. $$

The question itself tells us that $$S = k(10)^9$$, so from the last displayed line we can read off

$$ k \;=\; \sum_{r=0}^{9} (r+1)\Bigl(\tfrac{11}{10}\Bigr)^{r}. $$

Thus we only need to evaluate the finite series

$$ k \;=\; \sum_{r=0}^{9} (r+1)q^{\,r}\qquad\text{with}\quad q=\frac{11}{10}. $$

To do this, we first recall the standard formula for the sum of a geometric progression:

$$ \sum_{r=0}^{n} q^{\,r} \;=\; \frac{1-q^{\,n+1}}{1-q}\quad\text{for}\;q\ne1. $$

Next, to handle the factor $$r+1$$ in the general term, we differentiate this geometric sum with respect to $$q$$ and then manipulate it. Let us define

$$ G(q) \;=\; \sum_{r=0}^{9} q^{\,r}. $$

Using the formula above with $$n=9$$, we have

$$ G(q) \;=\; \frac{1-q^{\,10}}{1-q}. $$

Differentiate $$G(q)$$ with respect to $$q$$:

$$ G'(q) \;=\; \frac{d}{dq}\Bigl[\sum_{r=0}^{9} q^{\,r}\Bigr] \;=\; \sum_{r=0}^{9} r\,q^{\,r-1}. $$

Multiplying both sides by $$q$$ gives

$$ q\,G'(q) \;=\; \sum_{r=0}^{9} r\,q^{\,r}. $$

The series we actually need involves $$r+1$$, not just $$r$$. We therefore add $$G(q)$$ to $$qG'(q)$$:

$$ G(q) \;+\; q\,G'(q) \;=\; \sum_{r=0}^{9} q^{\,r} \;+\; \sum_{r=0}^{9} r\,q^{\,r} = \sum_{r=0}^{9} (r+1)\,q^{\,r}. $$

But this right-hand side is exactly the series for $$k$$. Hence

$$ k \;=\; G(q) \;+\; q\,G'(q). $$

Now we compute each part explicitly for $$q=\dfrac{11}{10}$$.

First compute $$G(q)$$:

$$ G(q) \;=\; \frac{1-q^{\,10}}{1-q}. $$

For $$q=\dfrac{11}{10}$$, we have $$1-q = 1-\dfrac{11}{10} = -\dfrac{1}{10}$$, so

$$ G\!\Bigl(\tfrac{11}{10}\Bigr) = \frac{1-\bigl(\tfrac{11}{10}\bigr)^{10}}{-\tfrac{1}{10}} = -10\Bigl(1-\bigl(\tfrac{11}{10}\bigr)^{10}\Bigr) = 10\Bigl(\bigl(\tfrac{11}{10}\bigr)^{10}-1\Bigr). $$

Next, we need $$G'(q)$$. Starting from

$$ G(q) = \frac{1-q^{\,10}}{1-q}, $$

we apply the quotient rule. Writing $$u = 1-q^{\,10}$$ and $$v = 1-q$$, the derivative is

$$ G'(q) \;=\; \frac{u'v - uv'}{v^{\,2}}. $$

We have $$u' = -10q^{\,9}$$ and $$v' = -1$$, so

$$ G'(q) \;=\; \frac{\bigl(-10q^{\,9}\bigr)(1-q) - (1-q^{\,10})(-1)}{(1-q)^{2}} = \frac{-10q^{\,9}(1-q) + 1 - q^{\,10}}{(1-q)^{2}}. $$

Multiplying this derivative by $$q$$ gives

$$ q\,G'(q) = \frac{-10q^{\,10}(1-q) + q(1-q^{\,10})}{(1-q)^{2}} = \frac{-10q^{\,10}(1-q) + q - q^{\,11}}{(1-q)^{2}}. $$

Now add $$G(q)$$ and $$qG'(q)$$. Using the common denominator $$(1-q)^{2}$$ we get

$$ k = G(q) + q\,G'(q) = \frac{(1-q^{\,10})(1-q) + \bigl[-10q^{\,10}(1-q) + q - q^{\,11}\bigr]}{(1-q)^{2}}. $$

Simplify the numerator carefully:

$$ (1-q^{\,10})(1-q) = (1-q^{\,10}) - (1-q^{\,10})q = 1 - q - q^{\,10} + q^{\,11}. $$

Adding the second bracket:

$$ \bigl[1 - q - q^{\,10} + q^{\,11}\bigr] \;+\; \bigl[-10q^{\,10}(1-q) + q - q^{\,11}\bigr]. $$

First expand $$-10q^{\,10}(1-q) = -10q^{\,10} + 10q^{\,11}$$. The whole numerator is

$$ 1 - q - q^{\,10} + q^{\,11} - 10q^{\,10} + 10q^{\,11} + q - q^{\,11} = 1 \;+\; (-q + q)\;+\;(-q^{\,10}-10q^{\,10}) \;+\;(q^{\,11}+10q^{\,11}-q^{\,11}). $$

The $$-q + q$$ terms cancel, and $$q^{\,11}+10q^{\,11}-q^{\,11}=10q^{\,11}$$, so the numerator reduces to

$$ 1 - 11q^{\,10} + 10q^{\,11}. $$

Therefore

$$ k = \frac{1 - 11q^{\,10} + 10q^{\,11}}{(1-q)^{2}}. $$

Because $$1-q = -\dfrac{1}{10}$$, we have $$(1-q)^{2} = \bigl(-\dfrac{1}{10}\bigr)^{2} = \dfrac{1}{100}$$, so dividing by $$(1-q)^{2}$$ is the same as multiplying by $$100$$:

$$ k = 100\bigl(1 - 11q^{\,10} + 10q^{\,11}\bigr). $$

Now substitute $$q=\dfrac{11}{10}$$:

$$ q^{\,10} = \Bigl(\frac{11}{10}\Bigr)^{10}, \qquad q^{\,11} = \Bigl(\frac{11}{10}\Bigr)^{11}. $$

Notice that

$$ 10q^{\,11} = 10\Bigl(\frac{11}{10}\Bigr)^{11} = 10\cdot\frac{11^{11}}{10^{11}} = \frac{11^{11}}{10^{10}} = 11\Bigl(\frac{11}{10}\Bigr)^{10} = 11q^{\,10}. $$

Hence in the expression $$1 - 11q^{\,10} + 10q^{\,11}$$, the last two terms precisely cancel:

$$ - 11q^{\,10} + 10q^{\,11} = - 11q^{\,10} + 11q^{\,10} = 0. $$

Therefore the entire bracket reduces simply to $$1$$, giving

$$ k = 100\times 1 = 100. $$

Hence, the correct answer is Option A.