If Σ_ =0^10 ( 10^ +1-1 10^ ) · ^11 C_ +1= α^11-11^11 10^10, then α is equal to :

$$\begin{aligned} & \sum_{\mathrm{r}=0}^{10}\left(\frac{1 \mathrm{r}^{\mathrm{r}-1}-1}{10^{\mathrm{r}}}\right){ }^{11} \mathrm{C}_{\mathrm{r}+1} \\ & =\sum_{\mathrm{r}=0}^{10}\left(10-\frac{1}{10^{\mathrm{r}}}\right)^{11} \mathrm{C}_{\mathrm{r}+1} \\ & =10 \sum_{\mathrm{r}=0}^{10} \mathrm{C}_{\mathrm{r}+1}-10 \sum\left({ }^{11} \mathrm{C}_{\mathrm{r}+1}\left(\frac{1}{10}\right)^{\mathrm{r}+1}\right) \\ & =10\left[{ }^{11} \mathrm{C}_1+{ }^{11} \mathrm{C}_2+\ldots . .+{ }^{11} \mathrm{C}_{11}\right] \\ & -10\left[{ }^{11} \mathrm{C}_1\left(\frac{1}{10}\right)^1+{ }^{11} \mathrm{C}_2\left(\frac{1}{10}\right)^2+\ldots . .+{ }^{11} \mathrm{C}_{11}\left(\frac{1}{10}\right)^{11}\right] \\ & =10\left[2^{11}-1\right]-10\left[\left(1+\frac{1}{10}\right)^{11}-1\right] \\ & =10(2)^{11}-10-\frac{11^{11}}{10^{10}}+10 \\ & =\frac{(20)^{11}-11^{11}}{10^{10}} \\ & \therefore \alpha=20\end{aligned}$$​r=0∑10​(10r1rr−1−1​)11Cr+1​=r=0∑10​(10−10r1​)11Cr+1​=10r=0∑10​Cr+1​−10∑(11Cr+1​(101​)r+1)=10[11C1​+11C2​+…..+11C11​]−10[11C1​(101​)1+11C2​(101​)2+…..+11C11​(101​)11]=10[211−1]−10[(1+101​)11−1]=10(2)11−10−10101111​+10=1010(20)11−1111​∴α=20​