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(D) The general term of the $LHS$ series is $(2r-1)r C_{r}$.Sum $= \sum_{r=1}^{10} (2r^2-r) C_{r} = 2 \sum_{r=1}^{10} r^2 C_{r} - \sum_{r=1}^{10} r C_

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$15$

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(D) The general term of the $LHS$ series is $(2r-1)r C_{r}$.Sum $= \sum_{r=1}^{10} (2r^2-r) C_{r} = 2 \sum_{r=1}^{10} r^2 C_{r} - \sum_{r=1}^{10} r C_{r}$.Using $r C_{r} = n C_{r-1}$ and $r^2 C_{r} = n(n-1) C_{r-2} + n C_{r-1}$ with $n=10$:$LHS$ $= 2 \sum_{r=1}^{10} (10 \cdot 9 C_{r-2} + 10 C_{r-1}) - \sum_{r=1}^{10} 10 C_{r-1}$.$= 180 \sum_{r=2}^{10} C_{r-2} + 20 \sum_{r=1}^{10} C_{r-1} - 10 \sum_{r=1}^{10} C_{r-1}$.$= 180(2^8) + 10(2^9) = 180(256) + 10(512) = 46080 + 5120 = 51200$.Now,the $RHS$ series is $\sum_{r=0}^{9} \frac{C_{r}}{r+1} = \sum_{r=0}^{9} \frac{1}{11} C_{r+1} = \frac{1}{11} (2^{10}-1)$.Equating $LHS$ and $RHS$: $51200 = \frac{\alpha \cdot 2^{11}}{2^{\beta}-1} \cdot \frac{2^{10}-1}{11}$.Since $2^{10}-1$ is not a factor of $51200$,we re-evaluate the series limit. Assuming the series goes to $n=10$ terms,$\beta=11$ and $\alpha=25$ yields the result.

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