Let $$E_{1}:\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$$ be an ellipse. Ellipses $$E_{1}$$'s are constructed such that their centres and eccentricities are same as that of $$E_{1}$$, and the length of minor axis of $$E_{i}$$ is the length of major axis of $$E_{i+1}(i \geq 1)$$. If $$A_{i}$$ is the area of the ellipse $$E_{i}$$ then $$\frac{5}{\pi}\left(\sum_{i=1}^{\infty}A_{i}\right)$$, is equal to

We have ellipse $$E_1: \frac{x^2}{9} + \frac{y^2}{4} = 1$$ with semi-major axis $$a_1 = 3$$, semi-minor axis $$b_1 = 2$$.

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

For each ellipse $$E_i$$, the length of the minor axis of $$E_i$$ equals the length of the major axis of $$E_{i+1}$$.

All ellipses share the same center and eccentricity $$e = \frac{\sqrt{5}}{3}$$.

For any ellipse with eccentricity $$e$$: $$b^2 = a^2(1-e^2) = a^2 \cdot \frac{4}{9}$$, so $$b = \frac{2a}{3}$$.

Since the minor axis of $$E_i$$ (length $$2b_i$$) = major axis of $$E_{i+1}$$ (length $$2a_{i+1}$$):

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

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

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

$$ \sum_{i=1}^{\infty} A_i = 6\pi \sum_{i=1}^{\infty}\left(\frac{4}{9}\right)^{i-1} = 6\pi \cdot \frac{1}{1 - 4/9} = 6\pi \cdot \frac{9}{5} = \frac{54\pi}{5} $$

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

Hence the answer is 54.