Series of squares of n integers - where is the mistake? Given the following two series:
$$1^3 + 2^3 + ... + n^3$$
$$0^3 + 1^3 + .... + (n-1)^3$$
I take the difference vertically of the two:
$$\left(1^3-0^3\right) + \left(2^3-1^3\right) + .... + \left(n^3-(n-1)^3\right)$$
This equals to $n^3$
If I now express this in sum notation:
$$\sum_{i=1}^n\left(i^3-(i-1)^3\right) = n^3$$
If I expand: $(i-1)^3 = i^3 - 3i^2 + 3i - 1$
Thus
$$\left(i^3 - (i-1)^3\right) = 3i^2 - 3i +1$$
My sum is now:
$$3\sum_{i=1}^n i^2 -3 \sum_{i=1}^n i + n = n^3$$
$$\sum_{i=1}^n i^2  = \frac{1}{3} \left(n^3 + 3 \frac{n(n-1)}{2}  -n\right)$$
And the expression on the RHS above is not $\frac{1}{6} n (n+1) (2n+1)$
I don't want to solve the above using forward difference, I want to keep it backward.

$$\sum_{i=1}^n i={n(n+1)\over2}\quad\text{ not }\quad {n(n-1)\over2}$$
