Let f ^ 1 (x) = (3x + 2)/(2x + 3), x ∈ R - \ - (3)/(2)\ . For n ≥ 2...

Given:

${f}^{1}\left(x\right) = \dfrac{3x + 2}{2x + 3}$

Now,

${f}^{2}\left(x\right) = {f}^{1}o{f}^{1}\left(x\right)$

$⇒ {f}^{2}\left(x\right) = \dfrac{3\left(\dfrac{3x + 2}{2x + 3}\right) + 2}{2\left(\dfrac{3x + 2}{2x + 3}\right) + 3}$

$⇒ {f}^{2}\left(x\right) = \dfrac{9x + 6 + 4x + 6}{6x + 4 + 6x + 9}$

$⇒ {f}^{2} ( x ) = \dfrac{13x + 12}{12x + 13}$

$⇒ {f}^{3}\left(x\right) = \dfrac{3\left(\dfrac{13x + 12}{12x + 13}\right) + 2}{2\left(\dfrac{13x + 12}{12x + 13}\right) + 3}$

$⇒ {f}^{3}\left(x\right) = \dfrac{63x + 62}{62x + 63}$

$⇒ {f}^{4}\left(x\right) = \dfrac{3\left(\dfrac{63x + 62}{62x + 63}\right) + 2}{2\left(\dfrac{63x + 62}{62x + 63}\right) + 3}$

$⇒ {f}^{4}\left(x\right) = \dfrac{313x + 312}{312x + 313}$

$∴ {f}^{5}\left(x\right) = \dfrac{1563x + 1562}{1562x + 1563} = \dfrac{ax + b}{bx + a}$

$∴ a + b = 1563 + 1562 = 3125$