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

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100

110

$\frac{121}{10}$

$\frac{441}{100}$

answer is A.

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Detailed Solution

The given series can be written as $k = 1 + 2 (\frac{11}{10}) + 3 {(\frac{11}{10})}^{2} + \dots + 9 {(\frac{11}{10})}^{8} + 10 {(\frac{11}{10})}^{9}$ …….(i)

On multiplying both sides by $(\frac{11}{10}),$ then

$\frac{11 k}{10} = (\frac{11}{10}) + 2 {(\frac{11}{10})}^{2} + 3 {(\frac{11}{10})}^{3} + \dots + 9 {(\frac{11}{10})}^{9} + 10 {(\frac{11}{10})}^{10} \dots$ (ii)

Now, on subtracting Eq. (ii) from Eq. (i), then

$\begin{array}{l} - \frac{k}{10} = \underset{10 tmes}{\underset{⏟}{1 + (\frac{11}{10}) + {(\frac{11}{10})}^{2} + \dots + {(\frac{11}{10})}^{2}}} - 10 {(\frac{11}{10})}^{10} \\ = \frac{1 \cdot \{{(\frac{11}{10})}^{10} - 1\}}{(\frac{11}{10} - 1)} - 10 {(\frac{11}{10})}^{10} \\ \Rightarrow k = - 100 \cdot \{{(\frac{11}{10})}^{10} - 1\} + 100 {(\frac{11}{10})}^{10} = 100 \end{array}$