Define the collections $\{E_1, E_2, E_3, \ldots\}$ of ellipses and $\{R_1, R_2, R_3, \ldots\}$ of rectangles as follows : $E_1 : \frac{x^2}{9} + \frac{y^2}{4} = 1$ ; $R_1$ : rectangle of largest area, with sides parallel to the axes, inscribed in $E_1$; $E_n$ : ellipse $\frac{x^2}{a_n^2} + \frac{y^2}{b_n^2} = 1$ of largest area inscribed in $R_{n-1}$, $n > 1$; $R_n$ : rectangle of largest area, with sides parallel to the axes, inscribed in $E_n, n > 1$. Then which of the following options is/are correct?

Define the collections {E₁, E₂, E₃, …} of ellipses and {R₁, R₂, R₃, …} of rectangles as follows : E₁ : x²/9 + y²/4 = 1 ; R₁ : rectangle of largest area, with sides parallel to the axes, inscribed in E₁; Eₙ : ellipse x²/aₙ² + y²/bₙ² = 1 of largest area inscribed in Rₙ₋₁, n > 1; Rₙ : rectangle of largest area, with sides parallel to the axes, inscribed in Eₙ, n > 1. Then which of the following options is/are correct? $E_1: \dfrac{x^2}{9} + \dfrac{y^2}{4} = 1$ $R_1 = \text{largest rectangle in } E_1;\quad E_n \text{ in } R_{n-1},\ R_n \text{ in } E_n$ $\text{Which of (1) - (4) is/are correct?}$ $E_1: a_1 = 3,\quad b_1 = 2$ $\text{max rectangle in } \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1:\ \text{area } 2ab,\ \text{corners } \left(\pm\dfrac{a}{\sqrt2},\pm\dfrac{b}{\sqrt2}\right)$ $\text{max ellipse in } W\times H \text{ rectangle}:\quad a = \dfrac{W}{2},\quad b = \dfrac{H}{2}$ $W_1 = 2\cdot\dfrac{a_1}{\sqrt2} = 3\sqrt2,\quad H_1 = 2\cdot\dfrac{b_1}{\sqrt2} = 2\sqrt2$ $\text{area } R_1 = 2a_1b_1 = 12$ $W_{n+1} = \dfrac{W_n}{\sqrt2},\qquad H_{n+1} = \dfrac{H_n}{\sqrt2}$ $W_n = 3\cdot 2^{\frac{2-n}{2}},\qquad H_n = 2\cdot 2^{\frac{2-n}{2}}$ $\dfrac{b_n}{a_n} = \dfrac{H_{n-1}}{W_{n-1}} = \dfrac{2}{3}\ \forall n \Rightarrow e = \sqrt{1-\dfrac49} = \dfrac{\sqrt5}{3}$ $\Rightarrow (1) \text{ FALSE}$ $a_9 = \dfrac{3}{16},\quad b_9 = \dfrac{1}{8}$ $\text{latus rectum} = \dfrac{2b_9^2}{a_9} = \dfrac{2/64}{3/16} = \dfrac16 \Rightarrow (2) \text{ TRUE}$ $\text{area } R_n = W_n H_n = 6\cdot 2^{2-n}$ $\sum_{n=1}^{\infty} 6\cdot 2^{2-n} = 6\cdot 4 = 24 \Rightarrow \text{each partial sum} < 24 \Rightarrow (3) \text{ TRUE}$ $c = \sqrt{a_9^2 - b_9^2} = \sqrt{\dfrac{9}{256} - \dfrac{1}{64}} = \dfrac{\sqrt5}{16} \ne \dfrac{\sqrt5}{32}$ $\Rightarrow (4) \text{ FALSE}$ $\boxed{(2)\ \text{and}\ (3)\ \text{are correct}}$