JEE MAIN - Mathematics (2025 - 2nd April Evening Shift - No. 17)
$$If\,\sum\limits_{r = 0}^{10} {({{{{10}^{r + 1}} - 1} \over {{{10}^r}}}).{}^{11}{C_{r + 1}} = {{{}_\alpha 11 - {{11}^{11}}} \over {{{10}^{10}}}},\,then\,\,\alpha \,\,is\,\,equal\,\,to:} $$
11
20
24
15
Explanation
$\begin{aligned} & \sum_{r=0}^{10}\left(\frac{10^{r-1}-1}{10^r}\right){ }^{11} \mathrm{C}_{\mathrm{r}+1} \\ & =\sum_{\mathrm{r}=0}^{10}\left(10-\frac{1}{10^r}\right){ }^{11} \mathrm{C}_{\mathrm{r}+1} \\ & =10 \sum_{\mathrm{r}=0}{ }^{11} \mathrm{C}_{\mathrm{r}+1}-10 \sum\left({ }^{11} \mathrm{C}_{\mathrm{r}+1}\left(\frac{1}{10}\right)^{r+1}\right) \\ & =10\left[{ }^{11} \mathrm{C}_1+{ }^{11} \mathrm{C}_2+\ldots . .+{ }^{11} \mathrm{C}_{11}\right] \\ & -10\left[{ }^{11} \mathrm{C}_1\left(\frac{1}{10}\right)^1+{ }^{11} \mathrm{C}_2\left(\frac{1}{10}\right)^2+\ldots . .+{ }^{11} \mathrm{C}_{11}\left(\frac{1}{10}\right)^{11}\right] \\ & =10\left[2^{11}-1\right]-10\left[\left(1+\frac{1}{10}\right)^{11}-1\right] \\ & =10(2)^{11}-10-\frac{11^{11}}{10^{10}}+10 \\ & =\frac{(20)^{11}-11^{11}}{10^{10}} \\ & \therefore \alpha=20\end{aligned}$
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