$\sum _{\mathrm{r}=0}^{300} {\mathrm{a}}_{\mathrm{r}}{\mathrm{x}}^{\mathrm{r}}={\left(1+\mathrm{x}+{\mathrm{x}}^2+{\mathrm{x}}^3\right)}^{100}.\text{ If }\mathrm{a}=\sum _{\mathrm{r}=0}^{300} {\mathrm{a}}_{\mathrm{r}},\text{ then }\sum _{\mathrm{r}=0}^{300} {\mathrm{ra}}_{\mathrm{r}}$ is equal to

$\sum _{\mathrm{r}=0}^{300} {\mathrm{a}}_{\mathrm{r}}\times {\mathrm{x}}^{\mathrm{r}}={\left(1+\mathrm{x}+{\mathrm{x}}^2+{\mathrm{x}}^3\right)}^{100}$

Clearly, 'a_r' is the coefficient of x^r in the expansion of ${\left(1+\mathrm{x}+{\mathrm{x}}^2+{\mathrm{x}}^3\right)}^{100}$.

Replacing x by 1/x in the given equation, we get

$\sum _{\mathrm{r}=0}^{300} {\mathrm{a}}_{\mathrm{r}}{\left(\frac{1}{\mathrm{x}}\right)}^{\mathrm{r}}=\frac{1}{{\mathrm{x}}^{300}}{\left({\mathrm{x}}^3+{\mathrm{x}}^2+\mathrm{x}+1\right)}^{100}$

or $\sum _{\mathrm{r}=0}^{300} {\mathrm{a}}_{\mathrm{r}}{\mathrm{x}}^{300-\mathrm{r}}={\left(1+\mathrm{x}+{\mathrm{x}}^2+{\mathrm{x}}^3\right)}^{100}$

Here, a_r represents the coefficient of ${\mathrm{x}}^{300-\mathrm{r}} \mathrm{in} {\left(1+\mathrm{x}+{\mathrm{x}}^2+{\mathrm{x}}^3\right)}^{100}$.

Thus, ${\mathrm{a}}_{\mathrm{r}}={\mathrm{a}}_{300-\mathrm{r}}$

Let

 $\begin{array}{c}\mathrm{I}=\sum _{\mathrm{r}=0}^{300} \mathrm{r}\times {\mathrm{a}}_{\mathrm{r}}\\ =\sum _{\mathrm{r}=0}^{300} (300-\mathrm{r}){\mathrm{a}}_{300-\mathrm{r}}\\ =\sum _{\mathrm{r}=0}^{300} (300-\mathrm{r}){\mathrm{a}}_{\mathrm{r}}\\ =300\sum _{\mathrm{r}=0}^{300} {\mathrm{a}}_{\mathrm{r}}-\sum _{\mathrm{r}=0}^{300} {\mathrm{ra}}_{\mathrm{r}}\end{array}$

$\begin{array}{c}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}2\mathrm{I}=300\mathrm{a}\\ \text{ or }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{I}=150\mathrm{a}\end{array}$