Please remind me of how this technique works ... We had a high school mathematics teacher who taught us a cool technique that I've forgotten. It can be used, for example, for developing a formula for the sum of squares for the first "n" integers. You start by making a column for Sn, and then determine the differences until you get a constant. See the picture.

(sorry about the rotated picture)
How do you proceed from here to the formula?
 A: Every polynomial that takes integer values over the integers can be represented with respect to the binomial base as a linear combination with integer coefficients. In our case:
$$ n^2 = \color{blue}{2}\binom{n}{2}+\color{blue}{1}\binom{n}{1}+\color{blue}{0}\binom{n}{0} \tag{1}$$
And that leads to:
$$ \sum_{n=1}^{N}n^2 = 2\binom{N+1}{3}+1\binom{N+1}{2}+0\binom{N+1}{1} = \frac{N(N+1)(2N+1)}{6}.\tag{2} $$
The blue coefficients appearing in $(1)$ can be computed through the forward difference operator:
$$ \begin{array}{ccccccccc} \color{blue}{0} && 1 && 4 && 9 && 16 \\ &\color{blue}{1} && 3 && 5 && 7 && \\ && \color{blue}{2} && 2 && 2 &&& \end{array}\tag{3}$$
Another example, for $n^3$.
$$ \begin{array}{ccccccccc} \color{blue}{0} && 1 && 8 && 27 && 64 \\ &\color{blue}{1} && 7 && 19 && 37 && \\ && \color{blue}{6} && 12 && 18 &&& \\ &&& \color{blue}{6} && 6 \end{array}\tag{3bis}$$
Gives:
$$ n^3 = \color{blue}{6}\binom{n}{3}+\color{blue}{6}\binom{n}{2}+\color{blue}{1}\binom{n}{1}\tag{1bis} $$
hence:
$$ \sum_{n=1}^{N}n^3 = 6\binom{N+1}{4}+6\binom{N+1}{3}+1\binom{N+1}{2}=\left(\frac{N(N+1)}{2}\right)^2.\tag{2bis}$$
You may be also interested in knowing that our "magic blue numbers" just depend on Stirling numbers of the second kind.
A: *

*backgrounded on the fact that the difference of two consecutive squares is an odd number:


$(n+1)^2-n^2=2n+1$
the difference of two consecutive odd numbers is always $2$ thus the self-saying technique.
So when we proceed to find a square $n^2$ we sum up all odd numbers from 1 to $2n-1$ which is an arithmetic sequence of distance 2.
$n^2=\frac{n(2n)}{2}=\frac{\frac{(2n)(2n)}{2}}{2}=\frac{\frac{(2n-1)(2n)}{2}+n}{2}=\frac{\binom{2n}{2}+n}{2}$
Also we note that 
$n^2=\frac{\frac{(2n)(2n)}{2}}{2}=\frac{\frac{(2n+1)(2n)}{2}-n}{2}=\frac{\binom{2n+1}{2}-n}{2}$
Summing up the two sums of two formulas  $$\sum_n{\frac{\binom{2i}{2}+n}{2}}+\sum_n{\frac{\binom{2i+1}{2}-n}{2}}=\sum_n{\frac{\binom{2i}{2}+\binom{2i+1}{2}}{2}}=\sum_{2n+1}{\frac{\binom{i}{2}}{2}}=2\sum{i^2}$$ is just the sum of values in second column of pascal triangle (even and odd binomials) until $2n+1$, divided by 2.
    1
    1 1 
    1 2  |1 |
    1 3  |3 | 1
    1 4  |6 | 4  1
    1 5  |10| 10 5  1
    1 6  |15| 20 15 6 1
    . .   \  \
    . .    \  \
    . .     \  \
    1 7   21 \35\ .  .  .  . 

Summing up the second column will result in the next row/column situated pascal value, which is known to be $\binom{2n+2}{3}$
So ... $$2\sum_n{i^2}=\frac{\binom{2n+2}{3}}{2}=\frac{\frac{2n(2n+1)(2n+2)}{6}}{2}$$
