Given ellipse $E_1$

$E_1: \frac{x^2}{9}+\frac{y^2}{4}=1$

Hence its semi-major and semi-minor axes are

$a_1=3, \quad b_1=2$

and its eccentricity is

$e=\sqrt{1-\frac{b_1^2}{a_1^2}}=\sqrt{1-\frac{4}{9}}=\frac{\sqrt{5}}{3}.$

Condition for all ellipses $E_i$

All ellipses have the same centre and the same eccentricity as $E_1$.

So for every ellipse $E_i$ with semi-major axis $a_i$ and semi-minor axis $b_i$,

$e^2=1-\frac{b_i^2}{a_i^2}=\frac{5}{9}.$

Therefore,

$\frac{b_i^2}{a_i^2}=\frac{4}{9} \implies \frac{b_i}{a_i}=\frac{2}{3}.$

Thus,

$b_i=\frac{2}{3}a_i.$

Relation between consecutive ellipses

The length of minor axis of $E_i$ equals the length of major axis of $E_{i+1}$.

Length of minor axis of $E_i$ is $2b_i$. Length of major axis of $E_{i+1}$ is $2a_{i+1}$.

Hence,

$2b_i=2a_{i+1} \implies a_{i+1}=b_i.$

Using $b_i=\frac{2}{3}a_i$,

$a_{i+1}=\frac{2}{3}a_i.$

So $\{a_i\}$ is a geometric progression with

$a_1=3, \quad a_i=3\left(\frac{2}{3}\right)^{i-1}.$

Also,

$b_i=\frac{2}{3}a_i=2\left(\frac{2}{3}\right)^{i-1}.$

Area of $E_i$

Area of an ellipse is

$A_i=\pi a_i b_i.$

Therefore,

$A_i=\pi \cdot 3\left(\frac{2}{3}\right)^{i-1} \cdot 2\left(\frac{2}{3}\right)^{i-1}$ $A_i=6\pi \left(\frac{4}{9}\right)^{i-1}.$

Sum of all areas

$\sum_{i=1}^{\infty} A_i=6\pi \sum_{i=1}^{\infty}\left(\frac{4}{9}\right)^{i-1}.$

This is an infinite GP with first term $1$ and common ratio $\frac{4}{9}$.

So,

$\sum_{i=1}^{\infty}\left(\frac{4}{9}\right)^{i-1}=\frac{1}{1-\frac{4}{9}}=\frac{1}{\frac{5}{9}}=\frac{9}{5}.$

Hence,

$\sum_{i=1}^{\infty} A_i=6\pi \cdot \frac{9}{5}=\frac{54\pi}{5}.$

Required value

$\frac{5}{\pi}\left(\sum_{i=1}^{\infty} A_i\right)=\frac{5}{\pi}\cdot \frac{54\pi}{5}=54.$

Comparison with stored answer

Our derived answer is $54$, which matches the stored correct answer.