If $${({}^{30}{C_1})^2} + 2{({}^{30}{C_2})^2} + 3{({}^{30}{C_3})^2}\, + \,...\, + \,30{({}^{30}{C_{30}})^2} = {{\alpha 60!} \over {{{(30!)}^2}}}$$ then $\alpha$ is equal to :

$$ \begin{aligned} & \mathrm{S}=0 \cdot\left({ }^{30} \mathrm{C}_0\right)^2+1 \cdot\left({ }^{30} \mathrm{C}_1\right)^2+2 \cdot\left({ }^{30} \mathrm{C}_2\right)^2+\ldots \ldots+30 \cdot\left({ }^{30} \mathrm{C}_{30}\right)^2 \\\\ & \mathrm{S}=30 \cdot\left({ }^{30} \mathrm{C}_0\right)^2+29 \cdot\left({ }^{30} \mathrm{C}_1\right)^2+28 \cdot\left({ }^{30} \mathrm{C}_2\right)^2+\ldots . \cdot+0 \cdot\left({ }^{30} \mathrm{C}_0\right)^2 \\\\ & 2 \mathrm{~S}=30 \cdot\left({ }^{30} \mathrm{C}_0{ }^2+{ }^{30} \mathrm{C}_1{ }^2+\ldots \ldots . \cdot+{ }^{30} \mathrm{C}_{30}{ }^2\right) \\\\ & \mathrm{S}=15 \cdot{ }^{60} \mathrm{C}_{30}=15 \cdot \frac{60 !}{(30 !)^2} \\\\ & \frac{15 \cdot 10 !}{(30 !)^2}=\frac{\alpha \cdot 60 !}{(30 !)^2} \\\\ & \Rightarrow \alpha=15 \end{aligned} $$ <br/><br/><b>Other Method :</b> <br/><br/><p>Given,</p> <p>$$1\,.\,{\left( {{}^{30}{C_1}} \right)^2} + 2\,.\,{\left( {{}^{30}{C_2}} \right)^2} + 3\,.\,{\left( {{}^{30}{C_3}} \right)^2}\, + \,.....\, + \,30\,.\,{\left( {{}^{30}{C_{30}}} \right)^2}$$</p> <p>$= \sum\limits_{r = 1}^{30} {r\,.\,{{\left( {{}^{30}{C_r}} \right)}^2}}$</p> <p>$= \sum\limits_{r = 1}^{30} {r\,.\,{}^{30}{C_r}\,.\,{}^{30}{C_r}}$</p> <p>$$ = \sum\limits_{r = 1}^{30} {r\,.\,{{30} \over r}\,.\,{}^{29}{C_{r - 1}}\,.\,{}^{30}{C_r}} $$</p> <p>$= 30\sum\limits_{r = 1}^{30} {{}^{29}{C_{r - 1}}\,.\,{}^{30}{C_r}}$</p> <p>$= 30 \times$ (Coefficient of ${x^{29}}$ in ${(1 + x)^{59}}$)</p> <p>$= 30 \times \left( {{}^{59}{C_{29}}} \right)$</p> <p>$= 30 \times {{60} \over {30}} \times {}^{59}{C_{29}} \times {{30} \over {60}}$</p> <p>$= 30 \times {}^{60}{C_{30}} \times {{30} \over {60}}$</p> <p>$= 15 \times {{60!} \over {30!\,30!}}$</p> <p>$\therefore$ $\alpha = 15$</p>