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.
