Why do these two summations not equate? Why is the below equation false?
(1) $\sum_{i=0}^n i^2 =  \ (1/4)\sum_{i=0}^{2n} {i}^2 $
When If we let $\ j=2i $ , and substitute $\ i=(j/2)$ into the leftmost equation above, then:
(2) $\sum_{j=0}^{2n} {(j/2)}^2 $
Which gives the same sum as:
$\ (1/4)\sum_{i=0}^{2n} {i}^2$
Note: I know the equation on line (1) is wrong, by actually summing the adjacent summations, but I can't see why this would be so.

$$\begin{align}
\sum_{i=0}^n i^2&=0^2+1^2+2^2+3^2+4^2+\cdots +n^2\qquad\qquad\qquad\qquad\qquad  (1)\\
\sum_{j=0}^{2n} (j/2)^2&=\left(\frac 02\right)^2+\left(\frac 12\right)^2+\left(\frac 22\right)^2+\left(\frac 32\right)^2+\left(\frac 42\right)^2\cdots+\cdots +\left(\frac n2\right)^2+\cdots+\left(\frac {2n}2\right)^2\\
&=\frac 14 \left(0^2+1^2+2^2+3^2+4^2+\cdots+(2n)^2\right)\\
&=0^2+\left(\frac 12\right)^2+1+\left(\frac 32\right)^2+2^2+\left(\frac 52\right)^2+\cdots+\left(\frac {n-1}2\right)^2+n^2\qquad \qquad (2)\end{align}$$
You can see clearly that $(1)\neq (2)\\$ as well as identify the additional terms in $(2)$.
What you probably wanted was 
$$\sum_{i=0}^n i^2=\frac 14\sum_{i=0}^n (2i)^2$$

You might be interested to know that
$$\sum_{i=0}^{2n} i^2=\sum_{i=1}^{2n} i^2=\sum_{i=1}^n (2i-1)^2+(2i)^2$$
