Explanation

(b)
The general term in the expansion of
${(1 + 2x + 3x}^{2})^{10}$
$is\ \ \sum_{}^{}\frac{10!}{r!s!t!}1^{r}(2x)^{5}\left( 3x^{2} \right)^{t}$
$= \frac{10!}{r!s!t!}2^{s} \times 3^{t} \times x^{s + 2t}$
Where $r + s + t = 10$
We have to find $a_{1}ie,$ coefficient of $x$
For the coefficient of $x^{1}$, we must have
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ s + 2t = 1$
But $\ \ \ \ \ \ \ r + s + t = 10$
$\therefore s = 1 - 2t\ and\ r = 9 + t$
Where $0 \leq r,s,t \leq 10$
Now, $t = 0 \Longrightarrow s = 1,\ r = 9$
For other, values of $t$, we get negative value $s$. So, there is only one term containing $x$ and its coefficient is
$\ \ \ \ \ \ \ \frac{10!}{9!1!0!}2^{1} \times 3^{0} = 20$
Hence, $a_{1} = 20$
Alternate On differentiating given equation w. r. t. $x$, we get
$10(1 + 2x + {3x}^{2})^{9} = a_{1} + 2a_{2}\ x + ... + 20a_{20}x^{19}$
Put $x = 0$, we get
$\ 20 = a_{1}$