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Q. Let $a$ and $b$ be the coefficient of $x ^3$ in $\left(1+ x +2 x ^2+3 x ^3\right)^3$ and $\left(1+ x +2 x ^2+3 x ^3+4 x ^4\right)^3$, respectively then

Binomial Theorem

Solution:

$(1+z)^3 \text { where } z=x\left(1+2 x+3 x^2\right) $
$1+{ }^3 C_1 z+{ }^3 C_2 z^2+{ }^3 C_3 z^3$
$\text { coefficient of } x^3 \text { in }(1+z)^3$
${ }^3 C_1(3)+{ }^3 C_2(4)+{ }^3 C_3(1)=22 $
$\Rightarrow a=22$
$\text { now again }(1+y)^3$
$\text { where } y=x\left(1+2 x+3 x^2+4 x^3\right) $
$(1+y)^3=1+{ }^3 C_1 y+{ }^3 C_2 y^2+{ }^3 C_3 y^3 $
$\therefore \text { coefficient of } x^3 \text { is } $
${ }^3 C_1(3)+{ }^3 C_2(4)+{ }^3 C_3(1) $
$=9+12+1=22 $
$\Rightarrow b=22 $
$\text { Hence a }=b \Rightarrow a+b=44 $