Summation of binomial coefficients multiplied by their index

Question

$\sum_{r=0}^{300}a_rx^r = (1+x+x^2+x^3)^{100}.$ 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?

Answer 1

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 $\sum ra_r = 4^{99} \cdot 100 \cdot 6 = 150a$

Answer 2

Hint

$$f(x)=\sum_{r=0}^{300} a_rx^r=(1+x+x^2+x^3)^{100}$$ $$f'(x)=\sum_{r=0}^{300} ra_rx^{r-1}=100(1+x+x^2+x^3)^{99}(1+2x+3x^2)$$ $$f'(1)=\sum_{r=0}^{300} ra_r=600\cdot 4^{99}=150\cdot 4^{100}$$ We also have $$a=f(1)=\sum_{r=0}^{300} a_r=4^{100}$$