16.
The sum is given by
2 [ 1 - (1/3)^15 ] / [ 1 - 1/3 ] = 3 {rounded }
17. The total distance is given by
20 [ 1 - (2/5)^(n )] /[ 3/5 ] -10 where n is the nth bounce
So ....for the 5th bounce we have
t10 + 20 [ 1 - (2/5)^(5)] [ 3/5 ] = 22 + 124 / 125 ft
