The coefficient of x⁴ in the expansion of ${\left(1+\mathrm{x}+{\mathrm{x}}^2\right)}^{10}$ is ….

As it is  given ${\mathrm{C}}_{\mathrm{r}}{=}^{25}{\mathrm{C}}_{\mathrm{r}}, \mathrm{so}$

${\mathrm{C}}_0+5\cdot {\mathrm{C}}_1+9\cdot {\mathrm{C}}_2+\dots +\left(101\right)\cdot {\mathrm{C}}_{25}$

$\begin{array}{l}=\sum _{\mathrm{r}=0}^{25} {(4\mathrm{r}+1)}^{25}{\mathrm{C}}_{\mathrm{r}}=4\sum _{\mathrm{r}=0}^{25} {\mathrm{r}}^{25}{\mathrm{C}}_{\mathrm{r}}+\sum _{\mathrm{r}=0}^{25}{ }^{25}{\mathrm{C}}_{\mathrm{r}}\\ =4\sum _{\mathrm{r}=1}^{25} \mathrm{r}{\left(\frac{25}{\mathrm{r}}\right)}^{24}{\mathrm{C}}_{\mathrm{r}-1}+\sum _{\mathrm{r}=0}^{25}{ }^{25}{\mathrm{C}}_{\mathrm{r}}\left[{\text{ as }}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}={\frac{\mathrm{n}}{\mathrm{r}}}^{\mathrm{n}-1}{\mathrm{C}}_{\mathrm{r}-1}\right]\\ =100\sum _{\mathrm{r}=1}^{25}{ }^{24}{\mathrm{C}}_{\mathrm{r}-1}+\sum _{\mathrm{r}=0}^{25}{ }^{25}{\mathrm{C}}_{\mathrm{r}}\\ =100\left(2^4\right)+2^{25}\left[{\because }^{\mathrm{n}}{\mathrm{C}}_0{+}^{\mathrm{n}}{\mathrm{C}}_1{+}^{\mathrm{n}}{\mathrm{C}}_2+\dots {+}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}=2^{\mathrm{n}}\right]\\ =(50+1)2^{25}=\left(51\right)\cdot 2^{25}+2^{25}\cdot \mathrm{k}\text{ (given) }\end{array}$

So, k= 51

Hence answer is 51.