Let ${\left(1+x+x^2\right)}^{100}=\sum _{r=0}^{200} a_r{x}^r$ and $a=\sum _{r=0}^{200} a_r$ then value of $\sum _{r=1}^{200} \frac{r{a}_r}{25a}$ is- 

 

$a=\sum _{r=0}^{200} a_r=\sum _{r=0}^{200} a_r(1{)}^r=3^{100}$

Differentiating  $\left(1\right),$ we   get

$100{\left(1+x+x^2\right)}^{99}(1+2x)=\sum _{r=1}^{200} r{a}_r{x}^{r-1}$

put$x = 1$, to obtain

$\begin{array}{l}100\left(3{)}^{99}\right(3)=\sum _{r=1}^{200} r{a}_r\\ \Rightarrow \sum _{r=1}^{200} \frac{r{a}_r}{25a}=\frac{100a}{25a}=4\end{array}$