If the coefficient of x^15 in the expansion of [(a x^3+1~b x^1 / 3)^15] is equal to the co…

Question

If the coefficient of $x^{15}$ in the expansion of $$\left(\mathrm{a} x^{3}+\frac{1}{\mathrm{~b} x^{1 / 3}}\right)^{15}$$ is equal to the coefficient of $x^{-15}$ in the expansion of $$\left(a x^{1 / 3}-\frac{1}{b x^{3}}\right)^{15}$$, where $a$ and $b$ are positive real numbers, then for each such ordered pair $$(\mathrm{a}, \mathrm{b})$$ :

Accepted Answer

For $$\left( {a{x^3} + {1 \over {b{x^{{1 \over 3}}}}}} \right)$$

$${T_{r + 1}} = {}^{15}{C_r}{(a{x^3})^{15 - r}}{\left( {{1 \over {b{x^{{1 \over 3}}}}}} \right)^1}$$

$$\therefore$$ $${x^{15}} \to 3(15 - r) - {r \over 3} = 15$$

$$ \Rightarrow 30 = {{10r} \over 3} \Rightarrow r = 9$$

Similarly, for $${\left( {a{x^{{1 \over 3}}} - {1 \over {b{x^3}}}} \right)^{15}}$$

$${T_{r + 1}} = {}^{15}{C_r}{\left( {a{x^{{1 \over 3}}}} \right)^{15 - r}}{\left( { - {1 \over {b{x^3}}}} \right)^2}$$

$$\therefore$$ For $${x^{ - 15}} \to {{15 - r} \over 3} - 3r =  - 15 \Rightarrow r = 6$$

$$\therefore$$ $${}^{15}{C_9}{{{a^6}} \over {{b^9}}} = {}^{15}{C_6}{{{a^9}} \over {{b^6}}} \Rightarrow ab = 1$$

Solutions · 1