Define the collections $\left\{E_1,E_2,E_3,,,,\right\}$  of ellipses and $\left\{R_1,R_2,R_3,...\right\}$  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?

$E_1:\frac{x^2}{9}+\frac{y^2}{4}=1\left(\text{ ellipse }\right),e=\frac{\sqrt{5}}{3}{R}_1:\text{ Max. Rectangle inscribed having sides parallel to axes }$

$\begin{array}{l}\text{ For this }A=(3\cos \theta ,2\sin \theta ),\text{ where }\theta =\frac{\pi }{4}\text{ i.e }A=\left(\frac{3}{\sqrt{2}},\frac{3}{\sqrt{2}}\right)\\ \text{ Sides: }3\sqrt{2},2\sqrt{2}\text{ Area of }{R}_1=\left(3\sqrt{2}\right)\left(2\sqrt{2}\right)\\ \text{ If a.b are semi axes of ellipse then }a\sqrt{2},b\sqrt{2}\text{ are sides. }\end{array}$

$\begin{array}{l}\text{ For }{E}_2:\frac{x^2}{{\left(\frac{3}{\sqrt{2}}\right)}^2}+\frac{y^2}{{\left(\frac{2}{\sqrt{2}}\right)}^2}=1\text{ i.e. semi axes are }\frac{a}{\sqrt{2}},\frac{b}{\sqrt{2}}\text{ . }\\ \therefore E_9:\frac{x^2}{{\left(\frac{3}{(\sqrt{2}{)}^8}\right)}^2}+\frac{y^2}{{\left(\frac{2}{(\sqrt{2}{)}^8}\right)}^2}\Rightarrow \text{ distance from centre to foucs of }{E}_9=\mathrm{ae}\end{array}$

$\frac{3}{(\sqrt{2}{)}^8}\times \frac{\sqrt{5}}{3}=\frac{\sqrt{5}}{16} \text{ L.L. }\mathrm{R}=\frac{2b^2}{a}=\frac{2\times {\left(\frac{2}{(\sqrt{2}{)}^8}\right)}^2}{\frac{3}{(\sqrt{2}{)}^8}}=\frac{2\times \frac{4}{256}}{\frac{3}{16}}=\frac{1}{32}\times \frac{16}{3}=\frac{1}{6}$

$\begin{array}{l}\text{ Eccentricities for all ellipses are equal. }\\ \text{ Also sum of areas }=R_1+R_2+R_3\dots \dots \dots \infty =4\left[\text{ Areas sum in }{\mathrm{Q}}_1\right]\\ =4\left[\frac{3}{\sqrt{2}}\frac{2}{\sqrt{2}}+\frac{3}{(\sqrt{2}{)}^2}\frac{2}{(\sqrt{2}{)}^2}+\frac{3}{(\sqrt{2}{)}^3}\frac{2}{\left({\sqrt{2}}^3\right)}+\dots \dots \right]=\left(4\right)\left(6\right)\left[\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\dots \dots .\right]\\ =24\times \frac{\frac{1}{2}}{1-\frac{1}{2}}=24\therefore \sum _{n=1}^N\left(\text{ area of }{\mathrm{R}}_{\mathrm{n}}\right)<24\end{array}$