If the coefficient of x^7 in (ax-1bx^2)^13 and the coefficient of x^-5 in (ax+1bx^2)^13 ar…

Question

If the coefficient of ${x}^{7}$ in ${(ax-\frac{1}{b{x}^{2}})}^{13}$ and the coefficient of ${x}^{-5}$ in ${(ax+\frac{1}{b{x}^{2}})}^{13}$ are equal, then ${a}^{4}{b}^{4}$ is equal to :

Accepted Answer

The given expression is ${(ax-\frac{1}{b{x}^{2}})}^{13}$

So,

$\begin{gathered} {T}_{r+1} ={}^{13}{C}_{r}(ax{)}^{13-r}{(-\frac{1}{b{x}^{2}})}^{r} \ \ ={}^{13}{C}_{r}(a{)}^{13-r}{x}^{13-r-2r}(-1/b{)}^{r} \ \ ={}^{13}{C}_{r}(a{)}^{13-r}{(-\frac{1}{b})}^{r}{x}^{13-3r} \end{gathered}$

For coefficient of ${x}^{7}$ in ${(ax-\frac{1}{b{x}^{2}})}^{13}$

$\begin{gathered} \, 13-3r=7 \ \Rightarrow 3r=6\Rightarrow r=2 \ 	herefore {x}^{7}={}^{13}{C}_{2}\cdot (a{)}^{11}\cdot \frac{1}{{b}^{2}} \end{gathered}$

Again, the another expression is ${(ax+\frac{1}{b{x}^{2}})}^{13}$

So, ${T}_{n+1}={}^{13}{C}_{r}(ax{)}^{13-r}{(\frac{1}{b{x}^{2}})}^{r}={}^{13}{C}_{r}(a{)}^{13-r}{(\frac{1}{b})}^{r}{x}^{13-3r}$

For coefficient ${x}^{-5}$ in ${(ax+\frac{1}{b{x}^{2}})}^{13}$

$\begin{gathered} 13-3r=-5 \ \Rightarrow r=6 \end{gathered}$

So, coefficient of ${x}^{-5}={}^{13}{C}_{6}(a{)}^{7}\frac{1}{{b}^{6}}$

Now, according to the question,

${}^{13}{C}_{2}(a{)}^{11}\frac{1}{{b}^{2}}={}^{13}{C}_{6}(a{)}^{7}\frac{1}{{b}^{6}}$

$\begin{gathered} \Rightarrow {a}^{4}{b}^{4}=\frac{{}^{13}{C}_{6}}{{}^{13}{C}_{2}} \ 	herefore {a}^{4}{b}^{4}=22 \end{gathered}$

Solutions · 1