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

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

Professor PC Thomas Classes · Answer

(8) : We have, $$\begin{equation*}\left(2 x^{2}+3 x+4ight)^{10}=\sum_{r=0}^{20} a_{r} x^{r} 	ag{i} \end{equation*}$$ Replacing $x$ by $\frac{2}{x}$, we get\ $\left(\frac{8}{x^{2}}+\frac{6}{x}+4ight)^{10}=\sum_{r=0}^{20} a_{r}\left(\frac{2}{x}ight)^{r}$\ $\Rightarrow \quad \frac{2^{10}\left(2 x^{2}+3 x+4ight)^{10}}{x^{20}}=\sum_{r=0}^{20} a_{r} 2^{r} x^{-r}$\ $\Rightarrow \sum_{r=0}^{20} a_{r} x^{r}=\sum_{r=0}^{20} a_{r} 2^{r-10} x^{20-r} \quad$ [Using (i)]\ On comparing the coefficient of $x^{7}$ from both sides, we get, $a_{7}=a_{13} 2^{3} \Rightarrow \frac{a_{7}}{a_{13}}=8$\