Summation of binomial coefficients multiplied by their index 
$\sum_{r=0}^{300}a_rx^r = (1+x+x^2+x^3)^{100}.$ If $a= \sum^{300}_{r=0}a_r$ then $\sum_{r=0}^{300}ra_r =?  $
a) $300a$
b) $100a$
c) $150a$
d) $75 a$

Attempt: 
First of all if we substitute $x=1$, we get $a= 4^{100}$
Then, using formula of summation of geometric progression: 
$\sum_{r=0}^{300}a_rx^r= (1-x^4)^{100}(1-x)^{-100}$
Using generalised binomial theorem, coefficient of $x^r$ in $(1-x)^{-100}$ is $\dbinom{99+r}{r}$ and that of $x^{4r}$ in $(1-x^4)^{100}$ is $(-1)^{100-r}\dbinom{n}{r}$
What do I do after this? 

You have been given
$$\sum_{r=0}^{300}a_rx^r = (1+x+x^2+x^3)^{100}$$
substitute $x = 1 $ to get $a = \sum a_r = 4^{100}$
Differentiate once w.r.t $x$ to get 
$$\sum_{r=0}^{300}ra_rx^{r-1} = 100(1+x+x^2+x^3)^{99} (1+2x+3x^2)$$
again substitute $x = 1$ to get $\sum ra_r = 4^{99} \cdot 100  \cdot 6 = 150a$
