Question

Let ${\left(2x^2+3x+4\right)}^{10}=\sum _{r=0}^{20}{a}_r{x}^r$, then $\frac{a_7}{a_{13}}=$

Answer 1

Finding the value of $\frac{a_7}{a_{13}}$:

Given, ${\left(2x^2+3x+4\right)}^{10}=\sum _{r=0}^{20}{a}_r{x}^r\dots \dots \left(1\right)$

To obtain the symmetry, replace $x$ by $\frac{2}{x}$ in above equation we get,

$$\begin{gathered} {\left[2{\left(\frac{2}{x}\right)}^2+3\left(\frac{2}{x}\right)+4\right]}^{10}=\underset{r=0}{\overset{20}{\sum a_r\frac{2^r}{x^r}}} \\ \Rightarrow {\left[\frac{8}{x^2}+\frac{6}{x}+4\right]}^{10}=\sum _{r=0}^{20}{a}_r\times 2^r\left(\frac{1}{x^r}\right) \\ \Rightarrow \frac{2^{10}}{x^{20}}{\left[2x^2+3x+4\right]}^{10}=\sum _{r=0}^{20}{a}_r{2}^r\left(\frac{1}{x^r}\right) \\ \Rightarrow {\left[2x^2+3x+4\right]}^{10}=\sum _{r=0}^{20}{a}_r{2}^{\left(r-10\right)}{x}^{\left(20-r\right)}\dots \dots \left(2\right) \end{gathered}$$

Substituting equation $\left(1\right)$ in equation $\left(2\right)$, we get

$\sum _{r=0}^{20}{a}_r{x}^r=\sum _{r=0}^{20}{a}_r{2}^{\left(r-10\right)}{x}^{\left(20-r\right)}$

Put $r=7$ in LHS and $r=13$ on RHS to find the coefficients of $x^7$ on both sides and compare them we get,

$$\begin{gathered} a_7=a_{13}{\left(2\right)}^3 \\ \Rightarrow \frac{a_7}{a_{13}}=8 \end{gathered}$$

Hence, the value of the ratio $\frac{a_7}{a_{13}}=8$.