Mathematics - Binomial Theorem Question with Solution | TestHub

Given that ${\left(a+bx+c{x}^2\right)}^{10}=\sum _{i=10}^{20}{p}_i{x}^i, a, b, c\in \mathrm{\mathbb{N}}$

General term : $\frac{10!}{r_1! r_2! r_3!}{\left(a\right)}^{r_1}{\left(bx\right)}^{r_2}{\left(c{x}^2\right)}^{r_3}$

For coefficient. of $x:r_2+2r_3=1$

Let $\begin{array}{ccc}{r}_1& r_2& r_3\\ 9& 1& 0\end{array}$

$\therefore$ coefficient of $x=\frac{10!}{9!}{a}^9{b}^1=20$

$\Rightarrow a^9·b=2 ......\left(\mathrm{i}\right)$

Coefficient of $\Rightarrow x^2:\frac{10!}{8! 2! 0!}{a}^8·b^2+\frac{10!}{9! 0! 1!}·a^9·c=210$

$\Rightarrow 45a^8·b^2+10·a^9·c=210$

$\Rightarrow 9a^8{b}^2+2a^9·c=42 .......\left(\mathrm{ii}\right)$

As $a, b, c\in N$, so by hit and trial in equation $\left(\mathrm{i}\right) \& \left(\mathrm{ii}\right)$ we get, $a=1, b=2, c=3$

Hence, $2\left(a+b+c\right)=2\left(3+2+1\right)=12$