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

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Question

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

Vedclass · Accepted Answer

$20$

Vedclass · Answer

$\sum_{ r =0}^{10}\left(\frac{10^{ r -1}-1}{10^{ r }}ight){ }^{11} C _{ r +1}$
$=\sum_{ r =0}^{10}\left(10-\frac{1}{10^{ r }}ight)^{11} C _{ r +1}$
$=10 \sum_{ r =0}^{10}{ }^{11} C _{ r +1}-10 \sum\left({ }^{11} C _{ r +1}\left(\frac{1}{10}ight)^{ r +1}ight)$
$=10\left[{ }^{11} C _1+{ }^{11} C _2+\ldots . .+{ }^{11} C _{11}ight]$
$-10\left[{ }^{11} C _1\left(\frac{1}{10}ight)^1+{ }^{11} C _2\left(\frac{1}{10}ight)^2+\ldots . .+{ }^{11} C _{11}\left(\frac{1}{10}ight)^{11}ight]$
$=10\left[2^{11}-1ight]-10\left[\left(1+\frac{1}{10}ight)^{11}-1ight]$
$=10(2)^{11}-10-\frac{11^{11}}{10^{10}}+10$
$=\frac{(20)^{11}-11^{11}}{10^{10}}$
$	herefore \alpha=20$

જો $\sum_{ r =0}^{10}\left(\frac{10^{ r +1}-1}{10^{ r }}\right) \cdot{ }^{11} C _{ r +1}=\frac{\alpha^{11}-11^{11}}{10^{10}}$ હોય તો $\alpha$ ની કિમંત મેળવો.

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