Mathematics - Binomial Theorem Question with Solution | TestHub

${\sum }_{\mathrm{r}=1}^9\left(\frac{\mathrm{r}+3}{2^{\mathrm{r}}}\right)\cdot {}^9{\mathrm{C}}_{\mathrm{r}}=\alpha {\left(\frac{3}{2}\right)}^9-\beta ,\alpha ,\beta \in \mathrm{N}$ Now, $\begin{array}{rl}& \sum _{\mathrm{r}=1}^9\left(\frac{\mathrm{r}+3}{2^{\mathrm{r}}}\right)\cdot {}^9{\mathrm{C}}_{\mathrm{r}}=\sum _{\mathrm{r}=1}^9\left(\frac{\mathrm{r}}{2^{\mathrm{r}}}\right)\cdot {}^9{\mathrm{C}}_{\mathrm{r}}+\sum _{\mathrm{r}=1}^9\left(\frac{3}{2^{\mathrm{r}}}\right)\cdot {}^9{\mathrm{C}}_{\mathrm{r}}\\ & =\sum _{\mathrm{r}=1}^9\left(\frac{9}{2^{\mathrm{r}}}\right)\cdot {}^8{\mathrm{C}}_{\mathrm{r}-1}+3\sum _{\mathrm{r}=1}^9{}^9{\mathrm{C}}_{\mathrm{r}}{\left(\frac{1}{2}\right)}^{\mathrm{r}}\left[\mathrm{U}\sin \mathrm{g}\frac{{}^9{\mathrm{C}}_{\mathrm{r}}}{{}^8{\mathrm{C}}_{\mathrm{r}-1}}=\frac{9}{\mathrm{r}}\right]\\ & =\frac{9}{2}\sum _{\mathrm{r}=1}^9{}^8{\mathrm{C}}_{\mathrm{r}-1}{\left(\frac{1}{2}\right)}^{\mathrm{r}-1}+3\left(\sum _{\mathrm{r}=0}^9\left({}^9{\mathrm{C}}_{\mathrm{r}}{\left(\frac{1}{2}\right)}^{\mathrm{r}}\right)-1\right)\\ & =\frac{9}{2}{\left(1+\frac{1}{2}\right)}^8+3\left({\left(1+\frac{1}{2}\right)}^9-1\right)\end{array}$ $=\frac{9}{2}\cdot {\left(\frac{3}{2}\right)}^8+3{\left(\frac{3}{2}\right)}^9-3=6\cdot {\left(\frac{3}{2}\right)}^9-3$ Hence, $\alpha =6,\beta =3$ Thus $(\alpha +\beta {)}^2=81$