JEE Main 2023 — Binomial Theorem Question with Solution

We are given that ${\left(a+bx+c{x}^2\right)}^{10}={\sum }_{i=0}^{20}{p}_i{x}^i$, 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 $(a+bx+c{x}^2{)}^{10}$ as follows:

$$\sum _{k_1+k_2+k_3=10}\frac{10!}{k_1!k_2!k_3!}{a}^{k_1}(bx{)}^{k_2}(c{x}^2{)}^{k_3}$$

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,

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

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

$$k_2=2,k_1=8,k_3=0\text{and}{k}_2=0,k_1=9,k_3=1$$

So,

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

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

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:

$$45(1{)}^8(2{)}^2+10(1{)}^{9}c=210⟹180+10c=210⟹c=3$$

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 $\boxed{12}$.