Discussion Forum

If the coefficient of $x^{5}$  in the expansion  of   $\left( ax^{2}+\frac{1}{bx}\right)^{13}$ is equal to the coefficient of $x^{-5}$  in the expansion of   $\left( ax^{}-\frac{1}{bx^{2}}\right)^{13}$ , then ab=

Answer:

Explanation:

 The general  term in the expansion of   $\left( ax^{2}+\frac{1}{bx}\right)^{13}$

   $T_{ (p+1)}=^{13}C_{p}(ax^{2})^{13-p}\frac{1}{(bx)^{p}}$

    = $^{13}C_{p}\frac{a^{13-p}}{b^{p}}x^{26-3p}$

  For, $x^{5}$, p=7

 So, coefficient of $x^{5}$ in the expansion of

   $\left(ax^{2}+\frac{1}{bx}\right)^{13}$   is   $^{13}C_{7}\frac{a^{6}}{b^{7}}$     [$\because$  p=7]

 Similarly, the general term in the expansion of   $\left( ax^{}-\frac{1}{bx^{2}}\right)^{13}$

     $T_{q+1}=^{13}C_{q}(ax)^{13-q}\left(-\frac{1}{bx^{2}}\right)^{q}$

    $=^{13}C_{q}(-1)^{q}\frac{a^{13-q}}{b^{q}}x^{13-q}$

  for $x^{-5}, q=6$

 So, coefficient  of $x^{-5}$  in the expansion of   $\left( ax^{}-\frac{1}{bx^{2}}\right)^{13}$  is 

$^{13}C_{6}\frac{a^{7}}{b^{6}}$

 According to the question,

$^{13}C_{7}\frac{a^{6}}{b^{7}}=^{13}C_{6}\frac{a^{7}}{b^{6}}$

$\Rightarrow$   ab=1