If $$\ (1+3x+x^2)^{10}=\sum_{r=0}^{20}a_r x^r\ $$ Then then what is the least number except 1 which divides the following:$$\ \sum_{r=0}^{20}(3r+1)a_r\ $$
EDIT: i have put x=1 then it is something like this: $$\ 5^{10}=\sum_{r=0}^{20}a_r x^r\ $$
$$\ 5^{10}=\ a_0 x^0+a_1x^1+...+a_{20}x^{20}\ $$
Now, simplifying $$\ \sum_{r=0}^{20}(3r+1)a_r\ $$ we get:
$$\ \sum_{r=0}^{20}3ra_r+\sum_{r=0}^{20}a_r\ $$
$$\ 3 \sum_{r=0}^{20}ra_r+\ 5^{10}\ $$
After this I am unable to solve
Differentiating with respect to $x$ gives $$10\ (1+3x+x^2)^{9}(2x+3)=\sum_{r=0}^{20}ra_r x^{r-1}\ $$ Multiply this by $3$ and add to the original expression to get $$30\ (1+3x+x^2)^{9}(2x+3)+(1+3x+x^2)^{10}=\sum_{r=0}^{20}3ra_r x^{r-1}+\sum_{r=0}^{20}a_r x^r$$ $$=\sum_{r=0}^{20}a_rx^{r-1}(3r+x)$$ Since $x$ is an arbitrary number, its value doesn't matter. What happens if $x=1$?