If ${\left({ }^{30}{\mathrm{C}}_1\right)}^2+2{\left({ }^{30}{\mathrm{C}}_2\right)}^2+3{\left({ }^{30}{\mathrm{C}}_3\right)}^2+\dots \dots +30{\left({ }^{30}{\mathrm{C}}_{30}\right)}^2=\frac{\mathrm{\alpha } 60\text{ ! }}{(30\text{ ! }{)}^2}\text{ then }\mathrm{\alpha }$ is equal to

${\left({ }^{30}{\mathrm{C}}_1\right)}^2+2{\left({ }^{30}{\mathrm{C}}_2\right)}^2+3{\left({ }^{30}{\mathrm{C}}_3\right)}^2+\dots .+30{\left({ }^{30}{\mathrm{C}}_{30}\right)}^2=\alpha \cdot \frac{\lfloor 60}{(\underset{\_}{\mid 30}{)}^2}$
Wkt $\mathrm{a}\cdot {\mathrm{C}}_0^2+(\mathrm{a}+\mathrm{d}){\mathrm{C}}_1^2+(\mathrm{a}+2\mathrm{d}){\mathrm{C}}_2^2+\dots ..+(\mathrm{a}+\mathrm{nd}){\mathrm{C}}_{\mathrm{n}}^2=\frac{(2\mathrm{a}+\mathrm{nd})}{2}{\cdot }^{2\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}$
Now ${\left.0\cdot {\left({ }^{30}{\mathrm{C}}_0\right)}^2+1\cdot {\left({ }^{30}{\mathrm{C}}_1\right)}^2+2^{30}{\mathrm{C}}_2\right)}^2+\dots \dots ..+30\cdot {\left({ }^{30}{\mathrm{C}}_{30}\right)}^2-0\cdot {\left({ }^{30}{\mathrm{C}}_0\right)}^2=\alpha \cdot \frac{\underset{\_}{60}}{(\underset{\_}{30}{)}^2}$
$\begin{array}{l}\Rightarrow {\left(\frac{2\left(0\right)+30\left(1\right)}{2}\right)}^{60}{\mathrm{C}}_{30}-0=\mathrm{\alpha }\cdot \left({ }^{60}{\mathrm{C}}_{30}\right)\text{ Here }\mathrm{a}=0,\mathrm{d}=1\\ \Rightarrow 15{\cdot }^{60}{\mathrm{C}}_{30}=\mathrm{\alpha }{\cdot }^{60}{\mathrm{C}}_{30}\\ \Rightarrow \mathrm{\alpha }=15\end{array}$