Mathematics - Binomial Theorem Question with Solution | TestHub

$\begin{array}{rl}& \sum _{\mathrm{r}=0}^{10}\left(\frac{1{\mathrm{r}}^{\mathrm{r}-1}-1}{10^{\mathrm{r}}}\right){}^{11}{\mathrm{C}}_{\mathrm{r}+1}\\ & =\sum _{\mathrm{r}=0}^{10}{\left(10-\frac{1}{10^{\mathrm{r}}}\right)}^{11}{\mathrm{C}}_{\mathrm{r}+1}\\ & =10\sum _{\mathrm{r}=0}^{10}{\mathrm{C}}_{\mathrm{r}+1}-10\sum \left({}^{11}{\mathrm{C}}_{\mathrm{r}+1}{\left(\frac{1}{10}\right)}^{\mathrm{r}+1}\right)\\ & =10\left[{}^{11}{\mathrm{C}}_1+{}^{11}{\mathrm{C}}_2+\dots ..+{}^{11}{\mathrm{C}}_{11}\right]\\ & -10\left[{}^{11}{\mathrm{C}}_1{\left(\frac{1}{10}\right)}^1+{}^{11}{\mathrm{C}}_2{\left(\frac{1}{10}\right)}^2+\dots ..+{}^{11}{\mathrm{C}}_{11}{\left(\frac{1}{10}\right)}^{11}\right]\\ & =10\left[2^{11}-1\right]-10\left[{\left(1+\frac{1}{10}\right)}^{11}-1\right]\\ & =10(2{)}^{11}-10-\frac{11^{11}}{10^{10}}+10\\ & =\frac{(20{)}^{11}-11^{11}}{10^{10}}\\ & \therefore \alpha =20\end{array}$