Question:

**Step 1:** Rewrite the given equation by factoring out a common term. The given equation is: $$(10)^9 + 2(11)^1(10)^8 + 3(11)^2(10)^7 + \cdots + 10(11)^9 = k(10)^9$$ The left side can be expressed as a sum: $$ \sum_{r=1}^{10} r (11)^{r-1} (10)^{10-r} = k(10)^9 $$ To isolate $k$, we factor out $(10)^9$ from each term in the sum on the left side: $$ \sum_{r=1}^{10} r \frac{(11)^{r-1}}{(10)^{r-1}} (10)^{10-r} (10)^{r-1} = \sum_{r=1}^{10} r \left(\frac{11}{10}\right)^{r-1} (10)^9 $$ Thus, the equation becomes: $$ (10)^9 \sum_{r=1}^{10} r \left(\frac{11}{10}\right)^{r-1} = k(10)^9 $$ Dividing both sides by $(10)^9$ yields the expression for $k$: $$ k = \sum_{r=1}^{10} r \left(\frac{11}{10}\right)^{r-1} $$ **Step 2:** Evaluate the sum using the derivative of a geometric series. Consider the geometric series $S(x)$ defined as: $$ S(x) = \sum_{r=0}^{n} x^r = 1 + x + x^2 + \cdots + x^n $$ The derivative of this series with respect to $x$ is: $$ S'(x) = \frac{d}{dx} \left( \sum_{r=0}^{n} x^r \right) = \sum_{r=1}^{n} r x^{r-1} $$ The closed form for $S(x)$ is: $$ S(x) = \frac{x^{n+1}-1}{x-1} $$ For this problem, we have $n=10$ and $x = \frac{11}{10}$. Therefore, $k = S'(x) \Big|_{x=11/10}$. First, we find the derivative of $S(x)$ using the quotient rule: $$ S'(x) = \frac{d}{dx} \left( \frac{x^{n+1}-1}{x-1} \right) = \frac{(n+1)x^n(x-1) - (x^{n+1}-1)(1)}{(x-1)^2} $$ Substitute $n=10$ into the derivative formula: $$ S'(x) = \frac{11x^{10}(x-1) - (x^{11}-1)}{(x-1)^2} $$ Now, substitute $x = \frac{11}{10}$ into the expression for $S'(x)$. Note that $x-1 = \frac{11}{10} - 1 = \frac{1}{10}$. $$ k = \frac{11 \left(\frac{11}{10}\right)^{10} \left(\frac{1}{10}\right) - \left(\left(\frac{11}{10}\right)^{11}-1\right)}{\left(\frac{1}{10}\right)^2} $$ $$ k = \frac{\frac{11 \cdot 11^{10}}{10^{10}} \cdot \frac{1}{10} - \left(\frac{11^{11}}{10^{11}}-1\right)}{\frac{1}{100}} $$ $$ k = \frac{\frac{11^{11}}{10^{11}} - \frac{11^{11}}{10^{11}} + 1}{\frac{1}{100}} $$ $$ k = \frac{1}{\frac{1}{100}} $$ $$ k = 100 $$