If $(1+2x)(1+x+x^2)^{n}=\sum_{r=0}^{2n+1} a_rx^r $ and $(1+x+x^2)^{s}=\sum_{r=0}^{2s} b_rx^r$, then value of $\frac{\sum_{s=0}^{n}\sum_{r=0}^{2s} b_r}{\sum_{r=0}^{2n+1} \frac{a_r}{r+1}}$ will be:
(A) $2n+1$
(B) $\frac{2n+1}{2}$
(C)$\frac{n+1}{2}$
(D) $n+1$
I think finding the answer in terms of $n$ will be very lengthy task. So I thought of taking $n=1$ as all options give different value for $n=1$ but even if I take n=1 , I don't know what to do with $s$. Please suggest something.
$$\int_0^1(1+2x)(1+x+x^2)^{n}dx=\int_0^1\sum_{r=0}^{2n+1} a_rx^rdx = \sum_{r=0}^{2n+1} \frac{a_r}{r+1} $$
LHS integration: $$= \frac{1}{n+1}(3^{n+1} - 1)$$
Using $(1+x+x^2)^{s}=\sum_{r=0}^{2s} b_rx^r$
Let x = 1, $$3^s = \sum_{r=0}^{2s} b_r$$ $$\sum_{s=0}^n \sum_{r=0}^{2s} b_r = \sum_{s=0}^n 3^s = \frac{3^{n+1} - 1}{2}$$
Therefore, $$\frac{\sum_{s=0}^{n}\sum_{r=0}^{2s} b_r}{\sum_{r=0}^{2n+1} \frac{a_r}{r+1}} = \frac{\frac{3^{n+1} - 1}{2}}{\frac{1}{n+1}(3^{n+1} - 1)} = \frac{n+1}{2}$$
