Question

Let $(2x^2+3x+4{)}^{10}=\sum _{r=0}^{20}{a}_r{x}^r.$ Then the value of $\frac{a_9}{a_{11}}$ is equal to

• A. $1$
• B. $2$
• C. $4$
• D. $\frac{1}{2}$

Answer 1

The correct option is B $2$

$(2x^2+3x+4{)}^{10}=\sum _{r=0}^{20}{a}_r{x}^r$
Replace $x$ by $\frac{2}{x},$ we get
${(\frac{8}{x^2}+\frac{6}{x}+4)}^{10}=\sum _{r=0}^{20}\frac{a_r\cdot 2^r}{x^r}$
$\Rightarrow \frac{2^{10}(4+3x+2x^2{)}^{10}}{x^{20}}=\sum _{r=0}^{20}{a}_r{2}^r{x}^{-r}$
$\Rightarrow 2^{10}\sum _{r=0}^{20}{a}_r{x}^r=\sum _{r=0}^{20}{a}_r{2}^r{x}^{20-r}$
Comparing coefficient of $x^9$ both sides, we get
$2^{10}\times a_9=a_{11}\cdot 2^{11}\Rightarrow \frac{a_9}{a_{11}}=2$