Question

Let $\left(2 x^{2}+3 x+4\right)^{10}= \displaystyle\sum_{r=0}^{20} a_{r} x^{r} .$ Then $\frac{a_{7}}{a_{13}}$ is equal to ______.

Answer 1

Given $\left(2 x^{2}+3 x+4\right)^{10}=\displaystyle\sum_{r=0}^{20} a_{r} x^{r} \ldots$ (1)
replace $x$ by $\frac{2}{x}$ in above identity :
$\frac{2^{10}\left(2 x ^{2}+3 x +4\right)^{10}}{ x ^{20}}$
$=\displaystyle\sum_{ r =0}^{20} \frac{ a _{ r } 2^{ r }}{ x ^{ r }}$
$\Rightarrow 2^{10} \displaystyle\sum_{r=0}^{20} a_{r} x^{r}$
$=\displaystyle\sum_{r=0}^{20} a_{r} 2^{r} x^{(20-r)}($ from (i))
now, comparing coefficient of $x^{7}$ from both sides
(take $r=7$ in L.H.S. $\&$ $r=13$ in R.H.S.
$2^{10} a _{7}= a _{13} 2^{13}$
$ \Rightarrow \frac{ a _{7}}{ a _{13}}=2^{3}=8$