JEE Main 2025 — 2 April, Evening Shift — Question 24

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

$$=(10-1)^{11} C_{1}+\left(\frac{10^{2}-1}{10}\right){ }^{11} C_{2}$$ $$+\ldots+\left(\frac{10^{11}-1}{10^{10}}\right){ }^{11} C_{11}$$ $$=10\left({ }^{11} C_{1}+{ }^{11} C_{2}+\ldots+{ }^{11} C_{11}\right)$$ $$-\left({ }^{11} C_{1}+\frac{{ }^{11} C_{2}}{10}+\ldots+\frac{{ }^{11} C_{11}}{10^{10}}\right)$$ $$=10\left(2^{11}-1\right)-10\left(\frac{{ }^{11} C_{1}}{10}+\frac{{ }^{11} C_{2}}{10^{2}}+\ldots+\frac{{ }^{11} C_{11}}{10^{11}}\right)$$ $$=10\left(2^{11}-1\right)-10\left(\left(1+\frac{1}{10}\right)^{11}-1\right)$$ $$=10\left(2^{11}-1\right)-10\left(\frac{11^{11}-10^{11}}{10^{11}}\right)$$

$=10\left(2^{11}-1\right)-\left(\frac{11^{11}-10^{11}}{10^{10}}\right)$

$=\frac{10^{11}\left(2^{11}-1\right)-11^{11}+10^{11}}{10^{10}}$

$=\frac{2^{11}\left(10^{11}\right)-11^{11}}{10^{10}}=\frac{20^{11}-11^{11}}{10^{10}}$

$\Rightarrow \alpha=20$