Question

Let ${\left(2x^2+3x+4\right)}^{10}=\sum _{r=0}^{20}{a}_r{x}^r.$ Then $\frac{a_7}{a_{13}}$ is equal to ______.

Answer 1

Answer: 8

Given ${\left(2x^2+3x+4\right)}^{10}=\sum _{r=0}^{20}{a}_r{x}^r\dots$ (1)
replace $x$ by $\frac{2}{x}$ in above identity :
$\frac{2^{10}{\left(2x^2+3x+4\right)}^{10}}{x^{20}}$
$=\sum _{r=0}^{20}\frac{a_r{2}^r}{x^r}$
$\Rightarrow 2^{10}\sum _{r=0}^{20}{a}_r{x}^r$
$=\sum _{r=0}^{20}{a}_r{2}^r{x}^{(20-r)}($ from (i))
now, comparing coefficient of $x^7$ from both sides
(take $r=7$ in L.H.S. $\&$ $r=13$ in R.H.S.
$2^{10}{a}_7=a_{13}{2}^{13}$
$\Rightarrow \frac{a_7}{a_{13}}=2^3=8$