If, _r=0^10(10^r+1-110^r).^11C_r+1=_ 11-11^1110^10,, then, , , , is, , equal, , to:

Question

$If\, \sum _{r=0}^{10}(\frac{{10}^{r+1}-1}{{10}^{r}}).{}^{11}{C}_{r+1}=\frac{{}_{\alpha }11-{11}^{11}}{{10}^{10}},\, then\, \, \alpha \, \, is\, \, equal\, \, to:$

Accepted Answer

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

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