Gregory from Magnus C of E School in Nottingham and Luke from St Patrick's School reasoned correctly. Here is Luke's explanation:

For this question I found a pattern in it:
For example C1 has length of 1 and C2 has length of
2
3


and there is a pattern in this because 1 x 2 = 2, so there is your numerator,
and to get your denominator you multiply 1 by 3, so there you have your
2
3

.

So to put this into easier words, it simply means multiply the numerator by 2 the whole way and multiply the denominator by 3.
So there is my strategy.

So The answer to C3 is
4
9


and the answer to C4 is
8
27


and finally C5 is
16
81

.

Cn is
æ
ç
è 		2
3
 ö
÷
ø 		n-1

 


and when n ® infinity, Cn ® 0

David from Gordonstoun School got the same result and added:
In a geometric progression,
a = 1st term
r = constant factor
n = number of terms

Any value in a geometric progression is arn-1
In this case
a = 1
r =
2
3

r =
2
3

(which is less than 1),
so the higher its power, the closer the result is to zero.
(Any positive number smaller than 1, to the power of infinity, tends to zero.
Therefore rn tends to zero)

So as n tends to infinity,
arn-1 tends to a ×0 = 0
So the Cantor Set's length is zero.

Liam from Wilbarston School reasoned in a similar way:

The length of Cn+1 is simply two thirds of the length of Cn,
as Cn+1 is purely Cn with the middle thirds removed.

Now taking Ln to be the length of Cn:
L2 =
2
3

,
L3 =
4
9

,
L4 =
8
27

etc. etc.

It's obvious that Ln =
æ
ç
è 		2
3
 ö
÷
ø 		n-1

 

.
So as n tends to infinity, Ln gets increasingly smaller, i.e. tends to zero.

Therefore the length of the Cantor set is zero. In fact, the Cantor set is a set of points, because endpoints of line segments will never be removed, only middle thirds.

And as Euclid said,

'A point is that which has no part'
i.e. a point has zero length, zero width and zero height.

Well done to you all.