The total distance is the sum of an infinite geometric series. We have #a=10# and #r=.8# in
#S = sum_{k=0}^{infty} a r^k #
It's a cliche, but we'll derive the formula one more time.
#rS = sum_{k=0}^{infty} r ar^k = sum_{k=0}^{infty} ar^{k+1} #
Now we substitute #k'=k+1# so #k=k'-1#
# rS = sum_{k'=1}^{infty} ar^{k'} #
It's just a dummy variable so we can drop the primes.
# rS = sum_{k=1}^{infty} ar^{k} #
#S = sum_{k=0}^{infty} a r^k = a + sum_{k=1}^{infty} a r^k#
#S - rS = a #
#S = a/{1-r}#
For our problem
#S = 10/(1 - .8) = 50 #
On which bounce is 85% means the remaining infinite part is 15%.
Let's call #S_n = sum_{k=0}^{n-1} ar^k#
We want to find the #n# such that #S_n = .85 S# or
# S - S_n = S - .85 S = .15 S #
#S-S_n# is the sum of the infinite series starting at the #n+1# st term. So it's a geometric series whose first term is #ar^n# instead of #a#. So
# S-S_n = b = {ar^n}/{1-r}#
#r^n = { b (1-r)}/a #
#n = ln( { b (1-r)}/a ) / ln r #
# b = .15 S = .15 (50) = 7.5 #
#n = ln( 7.5(1-.8)/10)/ln(.8) approx 8.5 #
If we want at least 85% traveled, we round up to
#n=9 # bounces
Check
# sum_{k=8}^{infty} 10(.8^k) = 10(.8^8)/(1-.8) =8.388608
#
#8.388608 / 50 = .168#
# sum_{k=9}^{infty} 10(.8^k) = 10(.8^9)/(1-.8) = 6.7108864 #
#6.7108864 /50 = 0.134#
That's where we pass the 15% threshold. #quad sqrt#
The value #a=10# is irrelevant.
