Let (a+bx+cx^2)^10= _i=0^20p_ix^i,a,b,c N.

Question

Let ${(a+bx+c{x}^{2})}^{10}=\sum _{i=0}^{20}{p}_{i}{x}^{i},a,b,c\in N$.

If ${p}_{1}=20$ and ${p}_{2}=210$, then

$2(a+b+c)$ is equal to :

Accepted Answer

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 $p_2 = 210$.

We need to find the value of $2(a+b+c)$.
Using the multinomial theorem, we can express the expansion of $(a + bx + cx^2)^{10}$ as follows:

$$
\sum\limits_{k_1+k_2+k_3=10} {{10!} \over {{k_1}!{k_2}!{k_3}!}} a^{k_1} (bx)^{k_2} (cx^2)^{k_3}
$$

Now we need to find the coefficients of $x^1$ and $x^2$ in the expansion:

For $x^1$ term, we have:

$k_2 = 1, k_1 = 9, k_3 = 0$

So,

$$
p_1 = {{10!} \over {9!1!0!}} a^9 b^1 = 10a^9 b
$$

For $x^2$ term, there are two possibilities:

$$
k_2 = 2, k_1 = 8, k_3 = 0 \quad \text{and} \quad k_2 = 0, k_1 = 9, k_3 = 1
$$

So,

$$
p_2 = {{10!} \over {8!2!0!}} a^8 b^2 + {{10!} \over {9!0!1!}} a^9 c = 45a^8 b^2 + 10a^9 c
$$

Now we are given $p_1 = 20$ and $p_2 = 210$. So,
$$
10a^9 b = 20 \implies a^9 b = 2
$$

and $45a^8 b^2 + 10a^9 c = 210$

Now, divide the second equation by $a^8$:
$45b^2 + 10ac = 210$

We know that $a^9 b = 2$. Taking the $9^{th}$ root of both sides:
$$
ab = \sqrt[9]{2}
$$

Now, let $k = ab = \sqrt[9]{2}$. We can rewrite the equation for $x^2$ term as:
$45k^2 + 10k^9 = 210$

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:

$a+b = 2$

and from the equation $a^9 b = 2$, we get $a = 1, b = 2$ or vice versa.

Now we need to find the value of $c$. We can use the equation for the $x^2$ term again:

$45a^8 b^2 + 10a^9 c = 210$

Using $a=1$ and $b=2$, we get:

$$
45(1)^8 (2)^2 + 10(1)^9 c = 210 \implies 180 + 10c = 210 \implies c = 3
$$

So, $a=1$, $b=2$, and $c=3$. Now, we need to find the value of $2(a+b+c)$:

$2(a+b+c) = 2(1+2+3) = 2(6) = 12$

Therefore, the answer is $\boxed{12}$.

Solutions · 1