If , then is equal to : [2025]

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Solution

Q.
If $\sum_{r = 0}^{10} (\frac{10^{r + 1} - 1}{10^{r}}) \cdot {}^{11}C_{r + 1} = \frac{α^{11} - 11^{11}}{10^{10}}$ , then $α$ is equal to : [2025]

1 11

2 15

3 20

4 24

Ans.
(3)

Consider $\frac{10^{r + 1} - 1}{10^{r}} = \frac{10^{r + 1}}{10^{r}} - \frac{1}{10^{r}} = 10 - \frac{1}{10^{r}}$

$∴ \sum_{r = 0}^{10} (\frac{10^{r + 1} - 1}{10^{r}}) {}^{11}C_{r + 1} = \sum_{r = 0}^{10} (10 - \frac{1}{10^{r}}) {}^{11}C_{r + 1}$

$= \sum_{r = 0}^{10} 10 {}^{11}C_{r + 1} - \sum_{r = 0}^{10} \frac{1}{10^{r}} {}^{11}C_{r + 1}$

Now, $\sum_{r = 0}^{10} 10 {}^{11}C_{r + 1} = 10 \sum_{r = 0}^{10} {}^{11}C_{r + 1} + {}^{11}C_{0} - {}^{11}C_{0}$

$= 10 (2^{11} - 1)$ ... (i)

Also, $\sum_{r = 0}^{10} \frac{1}{10^{r}} {}^{11}C_{r + 1} = \sum_{k = 1}^{11} \frac{1}{10^{k - 1}} {}^{11}C_{k} = 10 \sum_{k = 1}^{11} \frac{1}{10^{k}} {}^{11}C_{k}$ ... (ii)

Consider ${(1 + \frac{1}{10})}^{11} = \sum_{k = 0}^{11} {}^{11}C_{k} {(\frac{1}{10})}^{k}$

$= {}^{11}C_{0} + \sum_{k = 1}^{11} {}^{11}C_{k} {(\frac{1}{10})}^{k} = 1 + \sum_{k = 1}^{11} {}^{11}C_{k} {(\frac{1}{10})}^{k}$

$\Rightarrow \sum_{k = 1}^{11} {}^{11}C_{k} {(\frac{1}{10})}^{k} = {(\frac{11}{10})}^{11} - 1$

Substituting in (ii), we get

$\sum_{r = 0}^{10} {}^{11}C_{r + 1} {(\frac{1}{10})}^{r} = 10 ({(\frac{11}{10})}^{11} - 1)$ ... (iii)

On combining (i) and (iii), we get

$\sum_{r = 0}^{10} (\frac{10^{r + 1} - 1}{10^{r}}) {}^{11}C_{r + 1} = 10 (2^{11} - 1) - 10 ({(\frac{11}{10})}^{11} - 1)$

$= 10 \times 2^{11} - 10 - \frac{11^{11}}{10^{10}} + 10$

$= 10 \times 2^{11} - \frac{11^{11}}{10^{10}}$

$= \frac{{(20)}^{11} - 11^{11}}{10^{10}} \Rightarrow α = 20$ .

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