A ball is dropped from a height of 200 meters. After striking the floor it re-bounces to $$\frac{4}{5}th$$ of the height from where it fell. The total distance it travels before coming to rest is _____

For the 1st drop the distance travelled by the ball = 200m

After 1st drop it rebounces to a height of 4/5 * 200 and then falls from that height . The total distance travelled in this case would be $$2\cdot\frac{4}{5}\cdot200$$

Similarly the total distance travelled in next case would be $$2\cdot\frac{4}{5}\cdot\frac{4}{5}\cdot200$$

So total distance = 

200 + $$2\cdot200\cdot\left(\frac{4}{5}+\left(\frac{4}{5}\right)^2+\left(\frac{4}{5}\right)^{3\ }+....\right)$$

200+400$$\left(\frac{4}{\frac{5}{1-\frac{4}{5}}}\right)$$    = 1800m