Give a geometric proof that $\frac{2\cdot3}{4} + \frac{2\cdot3}{16} + \frac{2\cdot3}{64} + \cdots + \frac{2\cdot3}{4^n} + \cdots = 2$ I am suppose to give a "picture proof" for this but I am still having trouble proving it in the first place. I tried proof by induction but I do not think that is how I am suppose to answer the question. Any help would be greatly appreciated.
 A: I swiped this illustration from Wikipedia's article on geometric series: 

Perhaps it is suggestive?
A: 
Let S_0 be the rectangle with dimension(2x3);
S_n be the rectangle with dimension(2^(1-n),3x2^(-n));

A_required= Area coloured with pink, which is equivalent to:

           (2x3)/4+(2x3)/16+...    ;

A_complement= Area coloured with purple

            = Area complementary to A_required in S_0 /2;

We have:
A_required  = S_1 + S_2 + ...
A_complement= 2(S_2 + S_3 + ...)
So 
A_required - A_complement/2 = S_1 = 3/2 --(1)
A_required + A_complement = 3
A_required/2 + A_complement/2     = 3/2 --(2);
(1)+(2),
(3/2)A_required=3
Therefore A_required=2   
A: Hint:
Divide the equation
$$
\frac 64+\frac6{4^2}+\cdots = 2
$$
by $6$ to obtain 
$$
\frac 14+\frac 1{4^2}+\cdots = \frac 13
$$
For a geometrical proof, start with a unit square. Color one of the four quarters of the square. Then, repeat using one of the other quarters. 
Link to image:
http://3.bp.blogspot.com/_PnLYRqe0k9g/Smp9INIigTI/AAAAAAAAAI0/gQyMxj1hry0/s1600-h/Sum+Of+Fractions+Inverse+Powers+of+4.png
==Alternative ways to proof it:==
This can be done using geometric series.
Another way is to multiply by $4$ and rewrite it:
$$
\frac 14+\frac 1{4^2}+\cdots =S\\
4S = S + 1\\
S=\frac 13
$$
