Question

If $(1+2x+3x^2{)}^{10}=k_1+k_{2}x+\dots +k_{21}{x}^{20}$, then which of the following is/are correct?

• A. $k_2=20$
• B. $k_3=30$
• C. $k_3=210$
• D. $k_2=21$

Answer 1

Answer: (A), (C)

The correct options are
A $k_2=20$
C $k_3=210$

The general term in the expansion of $(1+2x+3x^2{)}^{10}$
$=\frac{10!}{(r!)(s!)(t!)}\left(1{)}^r\right(2x{)}^s(3x^2{)}^t=\frac{10!}{(r!)(s!)(t!)}(1{)}^r\left(2{)}^s\right(3{)}^t(x{)}^{s+2t}$

Where, $r+s+t=10$ and $0\le r,s,t\le 10$
$k_2$ is coefficient of $x$, so
$s+2t=1r+s+t=10$
As $r,s,t\in W$, so $t=0$ is only possiblity
$\Rightarrow s=1\Rightarrow r=9$
Now,
$k_2=\frac{10!(2{)}^1}{9!1!0!}=20$

$k_3$ is coefficient of $x^2$, so
$s+2t=2r+s+t=10$
So,
$r=8,s=2,t=0r=9,s=0,t=1$
Now,
$k_3=\frac{10!(2{)}^2}{8!2!0!}+\frac{10!(3{)}^1}{9!1!0!}\Rightarrow k_3=180+30=210$