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

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(D) Given $S_n = \frac{q^{n+1}-1}{q-1}$. The expression is $\sum_{r=1}^{101} {^{101}C_r} S_{r-1}$.Substituting $S_{r-1} = \frac{q^r-1}{q-1}$,we get $\

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$2^{100}$

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(D) Given $S_n = \frac{q^{n+1}-1}{q-1}$. The expression is $\sum_{r=1}^{101} {^{101}C_r} S_{r-1}$.Substituting $S_{r-1} = \frac{q^r-1}{q-1}$,we get $\sum_{r=1}^{101} {^{101}C_r} \frac{q^r-1}{q-1} = \frac{1}{q-1} \left( \sum_{r=1}^{101} {^{101}C_r} q^r - \sum_{r=1}^{101} {^{101}C_r} \right)$.Using the binomial theorem,$\sum_{r=1}^{101} {^{101}C_r} q^r = (1+q)^{101} - 1$ and $\sum_{r=1}^{101} {^{101}C_r} = 2^{101} - 1$.Thus,the expression becomes $\frac{1}{q-1} ((1+q)^{101} - 1 - (2^{101} - 1)) = \frac{(1+q)^{101} - 2^{101}}{q-1}$.Now,$T_{100} = \frac{(\frac{q+1}{2})^{101} - 1}{\frac{q+1}{2} - 1} = \frac{(\frac{q+1}{2})^{101} - 1}{\frac{q-1}{2}} = \frac{2((q+1)^{101} - 2^{101})}{2^{101}(q-1)} = \frac{(q+1)^{101} - 2^{101}}{2^{100}(q-1)}$.Comparing $\frac{(1+q)^{101} - 2^{101}}{q-1} = \alpha \cdot \frac{(1+q)^{101} - 2^{101}}{2^{100}(q-1)}$,we find $\alpha = 2^{100}$.

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