If ${\left(4{\mathrm{x}}^2+1\right)}^{\mathrm{n}}=\sum _{\mathrm{r}=0}^{\mathrm{n}} {\mathrm{a}}_{\mathrm{r}}{\left(1+{\mathrm{x}}^2\right)}^{\mathrm{n}-\mathrm{r}}{\mathrm{x}}^{2\mathrm{r}}$, then the value of $\sum _{\mathrm{r}=0}^{\mathrm{n}} {\mathrm{a}}_{\mathrm{r}}$ is

$\begin{array}{l}{\left(4{\mathrm{x}}^2+1\right)}^{\mathrm{n}}=\sum _{\mathrm{r}=0}^{\mathrm{n}} {\mathrm{a}}_{\mathrm{r}}{\left(1+{\mathrm{x}}^2\right)}^{\mathrm{n}-\mathrm{r}}{\mathrm{x}}^{2\mathrm{r}}\\ \sum _{\mathrm{r}=0}^{\mathrm{n}} {\mathrm{a}}_{\mathrm{r}}=sum of the coefficients={\left(4\right(1^2)+1)}^n= 5^n\\ \{to get sum of the coefficients put x = 1\}\end{array}$