If the coefficient of x ^ 7 in (ax - 1 b x ^ 2 ) ^ 13 and the...

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The coefficient of ${x}^{7}$ in the expansion of ${\left(a{x}^{2} - \dfrac{1}{\text{ bx }}\right)}$^{13}$is equal to the coefficient of${x}^{- 5}$in${\left(ax + \dfrac{1}{b{x}^{2}}\right)}^{13}$.</p><p>We know that, the general term${T}_{r + 1}$in the expansion${\left(a + b\right)}^{n}$is</p><p>${T}_{r + 1} = Cr n {a}^{n - r}{b}^{r}$</p><p>Applying to${\left(ax - \dfrac{1}{b{x}^{2}}\right)}^{13}$, we get</p><p>${T}_{r + 1} = Cr 13 {\left(ax\right)}^{13 - r}{\left(- \dfrac{1}{b{x}^{2}}\right)}^{r}$</p><p>$⇒ {T}_{r + 1} = {\left(- 1\right)}^{r} \times Cr 13 {\left(a\right)}^{13 - r}{\left(x\right)}^{13 - 3r}{\left(b\right)}^{- r}$</p><p>Therefore,$13 - 3r = 7 ⇒ r = 2$for coefficient of${x}^{7}$.</p><p>Thus,</p><p>${T}_{3} = C2 13 \left(\dfrac{{a}^{11}}{{b}^{2}}\right)$</p><p>Similarly, applying to${\left(ax + \dfrac{1}{b{x}^{2}}\right)}^{13}$, we get</p><p>${T}_{r + 1} = Cr 13 {\left(ax\right)}^{13 - r}{\left(\dfrac{1}{b{x}^{2}}\right)}^{r}$</p><p>$⇒ {T}_{r + 1} = Cr 13 {\left(a\right)}^{13 - r}{\left(x\right)}^{13 - 3r}{\left(b\right)}^{- r}$</p><p>Therefore,$13 - 3r = - 5$for coefficient of${x}^{- 5}$</p><p>$⇒ r = 6$</p><p>So,</p><p>${T}_{7} = C6 13 {\left(a\right)}^{7}{\left(b\right)}^{- 6}$</p><p>Hence, applying the given condition we get</p><p>$C2 13 \left(\dfrac{{a}^{11}}{{b}^{2}}\right) = C6 13 {\left(a\right)}^{7}{\left(b\right)}^{- 6}$</p><p>$⇒ {a}^{4}{b}^{4} = \dfrac{C6 13}{C2 13}$</p><p>$⇒ {a}^{4}{b}^{4} = \dfrac{13 !}{7 ! \cdot 6 !} \times \dfrac{2 ! \cdot 11 !}{13 !}$</p><p>$⇒ {a}^{4}{b}^{4} = \dfrac{11 \times 10 \times 9 \times 8}{6 \times 5 \times 4 \times 3}$</p><p>$⇒ {a}^{4}{b}^{4} = 22