The general term of the sum is \(T_r = r\,(11)^{r-1}(10)^{10-r}\) for \(r = 1, 2, \ldots, 10\), since \(T_1=(10)^9\) and \(T_{10}=10(11)^9\).

Factoring out \((10)^9\): \(T_r = (10)^9 \cdot r\,x^{r-1}\) where \(x = 11/10\), so the given sum equals \((10)^9\sum_{r=1}^{10} r\,x^{r-1} = K(10)^9\), giving \(K = \sum_{r=1}^{10} r\,x^{r-1}\).

Using the standard identity \(\sum_{r=1}^{n} r\,x^{r-1} = \dfrac{1-(n+1)x^n+n\,x^{n+1}}{(1-x)^2}\) with \(n=10\) and \(x=11/10\).

Since \(n\,x^{n+1} = 10\cdot\frac{11}{10}\cdot x^{10} = 11x^{10}\), the numerator becomes \(1-11x^{10}+11x^{10} = 1\).

The denominator is \((1-x)^2 = \left(1-\frac{11}{10}\right)^2 = \left(-\frac{1}{10}\right)^2 = \frac{1}{100}\).

So \(K = 1 \div \frac{1}{100} = 100\).