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Q)

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


( A ) $110$
( B ) $100$
( C ) $\frac{441}{100}$
( D ) $\frac{121}{10}$

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