If ${\left(1+2x+3x^2\right)}^{10}=a_0+a_{1}x+a_2{x}^2+\dots +a_{20}{x}^{20}$, then $a_1$ equals

The general term in the expansion of

${\left(1+2x+3x^2\right)}^{10}$ is

$\frac{10!}{r!s!t!}{1}^r(2x{)}^s{\left(3x^2\right)}^t$ where $r+s+t=10$

$=\frac{10!}{r!s!t!}{2}^s\times 3^t\times x^{s+2t}$

We have to find $a_1$ i.e. the 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 $0\le r,s,t\le 10$.

Now, $t=0\Rightarrow s=1, r=9$

For other values of$t$, we get negative values of $s.$ So, there is only one term containing $x$ and its coefficient is

$\frac{10!}{9!1!0!}\times 2^1\times 3^0=20$

Hence, $a_1=20$

ALITER     We have, 

$\begin{array}{l}{\left(1+2x+3x^2\right)}^{10}\\ {=}^{10}{C}_0{+}^{10}{C}_1\left(2x+3x^2\right)\\ {+}^{10}{C}_2{\left(2x+3x^2\right)}^2+\dots {+}^{10}{C}_{10}{\left(2x+3x^2\right)}^{10}\\ \therefore a_1=\text{ Coeff. of }x=20.\end{array}$