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

$(a)\;100\qquad(b)\;110\qquad(c)\;\large\frac{121}{10} \qquad(d)\;\large\frac{441}{100}$

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