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.
$\frac{i}{2}$ is an integer implies $j$ must be even, whereas your sum (2) is taken over integers $j$ regardless of odd or even.
The sum of the first n natural numbers is given by the formula sum1=n(n+1)(2n+1)/6. Consider the sum of the first 2n natural numbers. Let k=2n. Then by the above formula we have sum2=k(k+1)(2k+1)/6=(2n)(2n+1)(4n+1)/6 on substituting k=2n. Thus sum2 is not equal to (sum1)/4. It is always better to write the formula first and then checking that RHS=LHS. Hope it helps.
$$\begin{align} \sum_{i=0}^n i^2&=0^2+1^2+2^2+3^2+4^2+\cdots +n^2\qquad\qquad\qquad\qquad\qquad (1)\\ \sum_{j=0}^{2n} (j/2)^2&=\left(\frac 02\right)^2+\left(\frac 12\right)^2+\left(\frac 22\right)^2+\left(\frac 32\right)^2+\left(\frac 42\right)^2\cdots+\cdots +\left(\frac n2\right)^2+\cdots+\left(\frac {2n}2\right)^2\\ &=\frac 14 \left(0^2+1^2+2^2+3^2+4^2+\cdots+(2n)^2\right)\\ &=0^2+\left(\frac 12\right)^2+1+\left(\frac 32\right)^2+2^2+\left(\frac 52\right)^2+\cdots+\left(\frac {n-1}2\right)^2+n^2\qquad \qquad (2)\end{align}$$ 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 $$\sum_{i=0}^{2n} i^2=\sum_{i=1}^{2n} i^2=\sum_{i=1}^n (2i-1)^2+(2i)^2$$
