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7.Binomial Theorem
hard
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 :
A$15$
B$11$
C$24$
D$20$
(JEE MAIN-2025)
Solution
$\sum_{ r =0}^{10}\left(\frac{10^{ r -1}-1}{10^{ r }}\right){ }^{11} C _{ r +1}$
$=\sum_{ r =0}^{10}\left(10-\frac{1}{10^{ r }}\right)^{11} C _{ r +1}$
$=10 \sum_{ r =0}^{10}{ }^{11} C _{ r +1}-10 \sum\left({ }^{11} C _{ r +1}\left(\frac{1}{10}\right)^{ r +1}\right)$
$=10\left[{ }^{11} C _1+{ }^{11} C _2+\ldots . .+{ }^{11} C _{11}\right]$
$-10\left[{ }^{11} C _1\left(\frac{1}{10}\right)^1+{ }^{11} C _2\left(\frac{1}{10}\right)^2+\ldots . .+{ }^{11} C _{11}\left(\frac{1}{10}\right)^{11}\right]$
$=10\left[2^{11}-1\right]-10\left[\left(1+\frac{1}{10}\right)^{11}-1\right]$
$=10(2)^{11}-10-\frac{11^{11}}{10^{10}}+10$
$=\frac{(20)^{11}-11^{11}}{10^{10}}$
$\therefore \alpha=20$
$=\sum_{ r =0}^{10}\left(10-\frac{1}{10^{ r }}\right)^{11} C _{ r +1}$
$=10 \sum_{ r =0}^{10}{ }^{11} C _{ r +1}-10 \sum\left({ }^{11} C _{ r +1}\left(\frac{1}{10}\right)^{ r +1}\right)$
$=10\left[{ }^{11} C _1+{ }^{11} C _2+\ldots . .+{ }^{11} C _{11}\right]$
$-10\left[{ }^{11} C _1\left(\frac{1}{10}\right)^1+{ }^{11} C _2\left(\frac{1}{10}\right)^2+\ldots . .+{ }^{11} C _{11}\left(\frac{1}{10}\right)^{11}\right]$
$=10\left[2^{11}-1\right]-10\left[\left(1+\frac{1}{10}\right)^{11}-1\right]$
$=10(2)^{11}-10-\frac{11^{11}}{10^{10}}+10$
$=\frac{(20)^{11}-11^{11}}{10^{10}}$
$\therefore \alpha=20$
Standard 11
Mathematics
