If ${ }^{20} \mathrm{C}_{\mathrm{r}}$ is the co-efficient of $\mathrm{x}^{\mathrm{r}}$ in the expansion of $(1+x)^{20}$, then the value of $\sum_{r=0}^{20} r^{2}\,\,{ }^{20} C_{r}$ is equal to :

The expression $x^3 – 3x^2 – 9x + c$ can be written in the form $(x – a)^2 (x – b)$ if the values of $c$ is

If $\mathrm{b}$ is very small as compared to the value of $\mathrm{a}$, so that the cube and other higher powers of $\frac{b}{a}$ can be neglected in the identity $\frac{1}{a-b}+\frac{1}{a-2 b}+\frac{1}{a-3 b} \ldots .+\frac{1}{a-n b}=\alpha n+\beta n^{2}+\gamma n^{3}$, then the value of $\gamma$ is:

Let $\alpha=\sum_{k=0}^n\left(\frac{\left({ }^n C_k\right)^2}{k+1}\right)$ and $\beta=\sum_{k=0}^{n-1}\left(\frac{{ }^n C_k{ }^n C_{k+1}}{k+2}\right)$. If $5 \alpha=6 \beta$, then $n$ equals

$\left( {\left( {\begin{array}{*{20}{c}}
{21}\\
1
\end{array}} \right) – \left( {\begin{array}{*{20}{c}}
{10}\\
1
\end{array}} \right)} \right) + \left( {\left( {\begin{array}{*{20}{c}}
{21}\\
2
\end{array}} \right) – \left( {\begin{array}{*{20}{c}}
{10}\\
2
\end{array}} \right)} \right)$$ + \left( {\left( {\begin{array}{*{20}{c}}
{21}\\
3
\end{array}} \right) – \left( {\begin{array}{*{20}{c}}
{10}\\
3
\end{array}} \right)} \right) + \;.\;.\;.$$ + \left( {\left( {\begin{array}{*{20}{c}}
{21}\\
{10}
\end{array}} \right) – \left( {\begin{array}{*{20}{c}}
{10}\\
{10}
\end{array}} \right)} \right) = $