4. After the first fall from the height a the ball will go up by some height and down by the same height.
As follows: a, ra (up), ra (down), r*ra (up), r*ra (down) and so on.
We ought to compute the sum of this (infinite) series. Let's group the terms:
a + (r*a+r*a) + (r^2*a + r^2*a) + (r^3*a + r^3*a) + ... =
= a + a*2*(r + r^2 + r^3 + ...). Again we see a geometric progression, its ratio r<1 and the first term is also r. It is convergent to the r/(1-r).
So the entire sum equals to a + 2*a*r/(1-r) = a*(1 + 2/(1-r)) = a*(1+r)/(1-r).
Then at the first step we shade the area A0*4*(1/4)*(1/4) = A/4.And the square that remains for shading at the second step also has area A/4.
At the second step we shade 1/4 of A/4 and leave for the third step A/(4^2).
And so on, on the n-th step we shade the area A/(4^n).
It is the sum of an infinite geometric progression with the first term A/4 and the ratio 1/4. So this sum is equal to A*(1/4)/(1-1/4) = A*(1/4)/(3/4) = A*(1/3).
So the answer 1s 1/3.
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