Let S_n = 1 + q + q^2 + ..... + q^n and T_n = | vedclass

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(D) Given $S_n = \frac{q^{n+1} - 1}{q - 1}$.We need to evaluate $\sum_{r=1}^{101} {^{101}C_r} S_{r-1}$.$= \sum_{r=1}^{101} {^{101}C_r} \left( \frac{q^

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$2^{100}$

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(D) Given $S_n = \frac{q^{n+1} - 1}{q - 1}$.We need to evaluate $\sum_{r=1}^{101} {^{101}C_r} S_{r-1}$.$= \sum_{r=1}^{101} {^{101}C_r} \left( \frac{q^r - 1}{q - 1} \right)$$= \frac{1}{q - 1} \left( \sum_{r=1}^{101} {^{101}C_r} q^r - \sum_{r=1}^{101} {^{101}C_r} \right)$$= \frac{1}{q - 1} \left( ((1 + q)^{101} - 1) - (2^{101} - 1) \right)$$= \frac{(1 + q)^{101} - 2^{101}}{q - 1}$.Now,$T_{100} = \frac{(\frac{q+1}{2})^{101} - 1}{\frac{q+1}{2} - 1} = \frac{(\frac{q+1}{2})^{101} - 1}{\frac{q-1}{2}} = \frac{2}{q-1} \left( \frac{(q+1)^{101} - 2^{101}}{2^{101}} \right) = \frac{(q+1)^{101} - 2^{101}}{(q-1) 2^{100}}$.Given $\sum_{r=1}^{101} {^{101}C_r} S_{r-1} = \alpha T_{100}$,we have:$\frac{(1 + q)^{101} - 2^{101}}{q - 1} = \alpha \cdot \frac{(1 + q)^{101} - 2^{101}}{(q - 1) 2^{100}}$.Therefore,$\alpha = 2^{100}$.

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