If sum_{r=0}^{10} left( frac{10^{r+1}-1}{10^r | vedclass

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(D) Let $S = \sum_{r=0}^{10} \left( \frac{10^{r+1}-1}{10^r} \right) {}^{11}C_{r+1}$.$= \sum_{r=0}^{10} \left( 10 - \frac{1}{10^r} \right) {}^{11}C_{r+

Vedclass · Accepted Answer

$20$

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(D) Let $S = \sum_{r=0}^{10} \left( \frac{10^{r+1}-1}{10^r} \right) {}^{11}C_{r+1}$.$= \sum_{r=0}^{10} \left( 10 - \frac{1}{10^r} \right) {}^{11}C_{r+1}$.$= 10 \sum_{r=0}^{10} {}^{11}C_{r+1} - \sum_{r=0}^{10} {}^{11}C_{r+1} \left( \frac{1}{10^r} \right)$.$= 10 \sum_{r=0}^{10} {}^{11}C_{r+1} - 10 \sum_{r=0}^{10} {}^{11}C_{r+1} \left( \frac{1}{10^{r+1}} \right)$.$= 10 \left( {}^{11}C_1 + {}^{11}C_2 + \dots + {}^{11}C_{11} \right) - 10 \left( {}^{11}C_1 \left( \frac{1}{10} \right)^1 + {}^{11}C_2 \left( \frac{1}{10} \right)^2 + \dots + {}^{11}C_{11} \left( \frac{1}{10} \right)^{11} \right)$.Using the binomial expansion $(1+x)^n = \sum_{k=0}^n {}^{n}C_k x^k$,we have $\sum_{k=1}^{11} {}^{11}C_k = 2^{11}-1$ and $\sum_{k=1}^{11} {}^{11}C_k (\frac{1}{10})^k = (1+\frac{1}{10})^{11} - 1$.$S = 10(2^{11}-1) - 10((\frac{11}{10})^{11} - 1)$.$S = 10 \cdot 2^{11} - 10 - 10 \cdot \frac{11^{11}}{10^{11}} + 10$.$S = \frac{10 \cdot 2^{11} \cdot 10^{10} - 11^{11}}{10^{10}} = \frac{20^{11} - 11^{11}}{10^{10}}$.Comparing with $\frac{\alpha^{11}-11^{11}}{10^{10}}$,we get $\alpha = 20$.

If $\sum_{r=0}^{10} \left( \frac{10^{r+1}-1}{10^r} \right) \cdot {}^{11}C_{r+1} = \frac{\alpha^{11}-11^{11}}{10^{10}}$,then $\alpha$ is equal to :

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