Let $\left(a+b x+c x^{2}\right)^{10}=\sum\limits_{i=0}^{20} p_{i} x^{i}, a, b, c \in \mathbb{N}$.

If $p_{1}=20$ and $p_{2}=210$, then $2(a+b+c)$ is equal to :

<p>We are given that $\left(a+bx+cx^2\right)^{10} = \sum_{i=0}^{20} p_i x^i$, and we are given that $p_1 = 20$ and $p_2 = 210$.<br/><br/> We need to find the value of $2(a+b+c)$.</p> Using the multinomial theorem, we can express the expansion of $(a + bx + cx^2)^{10}$ as follows: <br/><br/>$$ \sum\limits_{k_1+k_2+k_3=10} {{10!} \over {{k_1}!{k_2}!{k_3}!}} a^{k_1} (bx)^{k_2} (cx^2)^{k_3} $$ <br/><br/>Now we need to find the coefficients of $x^1$ and $x^2$ in the expansion: <br/><br/>For $x^1$ term, we have: <br/><br/>$k_2 = 1, k_1 = 9, k_3 = 0$ <br/><br/>So, <br/><br/>$p_1 = {{10!} \over {9!1!0!}} a^9 b^1 = 10a^9 b$ <br/><br/>For $x^2$ term, there are two possibilities: <br/><br/>$k_2 = 2, k_1 = 8, k_3 = 0 \quad \text{and} \quad k_2 = 0, k_1 = 9, k_3 = 1$ <br/><br/>So, <br/><br/>$$ p_2 = {{10!} \over {8!2!0!}} a^8 b^2 + {{10!} \over {9!0!1!}} a^9 c = 45a^8 b^2 + 10a^9 c $$ <p>Now we are given $p_1 = 20$ and $p_2 = 210$. So, $10a^9 b = 20 \implies a^9 b = 2$</p> <p>and $45a^8 b^2 + 10a^9 c = 210$</p> <p>Now, divide the second equation by $a^8$: $45b^2 + 10ac = 210$</p> <p>We know that $a^9 b = 2$. Taking the $9^{th}$ root of both sides: $ab = \sqrt[9]{2}$</p> <p>Now, let $k = ab = \sqrt[9]{2}$. We can rewrite the equation for $x^2$ term as: $45k^2 + 10k^9 = 210$</p> <p>From the equation $ab = k = \sqrt[9]{2}$, we know that $a$ and $b$ are positive integers. Thus, $k = 2$ (as both $a$ and $b$ must be factors of 2). Now we have:</p> <p>$a+b = 2$</p> <p>and from the equation $a^9 b = 2$, we get $a = 1, b = 2$ or vice versa. </p> <p>Now we need to find the value of $c$. We can use the equation for the $x^2$ term again:</p> <p>$45a^8 b^2 + 10a^9 c = 210$</p> <p>Using $a=1$ and $b=2$, we get:</p> <p>$45(1)^8 (2)^2 + 10(1)^9 c = 210 \implies 180 + 10c = 210 \implies c = 3$</p> <p>So, $a=1$, $b=2$, and $c=3$. Now, we need to find the value of $2(a+b+c)$:</p> <p>$2(a+b+c) = 2(1+2+3) = 2(6) = 12$</p> <p>Therefore, the answer is $\boxed{12}$.</p>