Question

A frog sits at one end of a table which is 2m long. In its first jump the frog goes a distance of 1m along the table, with its second jump 1/2 m, with its third jump 1/4 m and so on. After how many jumps will the frog be within 1 cm of the far end of the table?

Answer 1

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
$S_n= \frac{a_1(1-r^n)}{1-r}$
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

$1.99= \frac{1(1- (\frac{1}{2})^n)}{1-\frac{1}{2}}$
$1.99= \frac{(1- (\frac{1}{2})^n)}{\frac{1}{2}}$
$1.99= 2(1- (\frac{1}{2})^n)$
divide both sides by 2
$\frac{1.99}{2} = 1- (\frac{1}{2})^n$
times -1
$\frac{-1.99}{2} = (\frac{1}{2})^n-1$
add 1 or 2/2 to both sides
$\frac{0.01}{2} = (\frac{1}{2})^n$
take the ln of both sides
$ln(\frac{0.01}{2}) = ln((\frac{1}{2})^n)$
$ln(\frac{0.01}{2}) = n ln(\frac{1}{2})$
divide both sides by ln(1/2)
$\frac{ln(\frac{0.01}{2})}{ln(\frac{1}{2})} =n$
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