ah, this is an infinite sum question, or a sum of geometric sequence
when will the sum reach 1cm of the edge or about 1.99m
so
2m=200cm
within 1cm means at least 1.99m
so
we will use m and not cm for consitancy
sum of geometric sequence is
[tex]S_n= \frac{a_1(1-r^n)}{1-r} [/tex]
a1=first term=initial jjump=1
r=common ratio=1/2
n=?, we ar solving for that
so
we want it to equal 1.99 so
[tex]1.99= \frac{1(1- (\frac{1}{2})^n)}{1-\frac{1}{2}} [/tex]
[tex]1.99= \frac{(1- (\frac{1}{2})^n)}{\frac{1}{2}} [/tex]
[tex]1.99= 2(1- (\frac{1}{2})^n)[/tex]
divide both sides by 2
[tex]\frac{1.99}{2} = 1- (\frac{1}{2})^n[/tex]
times -1
[tex]\frac{-1.99}{2} = (\frac{1}{2})^n-1[/tex]
add 1 or 2/2 to both sides
[tex]\frac{0.01}{2} = (\frac{1}{2})^n[/tex]
take the ln of both sides
[tex]ln(\frac{0.01}{2}) = ln((\frac{1}{2})^n)[/tex]
[tex]ln(\frac{0.01}{2}) = n ln(\frac{1}{2})[/tex]
divide both sides by ln(1/2)
[tex] \frac{ln(\frac{0.01}{2})}{ln(\frac{1}{2})} =n[/tex]
use your calculatro to find that n≈7.64386
so on 7th jump, it is not yet at 1cm to the edge but at 8th jump, it is past
so 8th jump
