<s>[INST] Proof of $n^3/3 + n^2/2 + n/6 = [1^2 + 2^2 + ... + n^2]$? I am having a hard time following this proof. Here is how it goes.
$$
(k+1)^3 = k^3+3k^2+3k+1\\
3k^2+3k+1 = (k+1)^3-k^3\\
$$
if $ k = 1, 2, 3, ... , n-1$ we add all the 5 formulas like this 
$$
3(1)^2+3(1)+1 = ((1)+1)^3-(1)^3\\
3(2)^2+3(2)+1 = ((2)+1)^3-(2)^3\\
3(3)^2+3(3)+1 = ((3)+1)^3-(3)^3\\
3(4)^2+3(4)+1 = ((4)+1)^3-(4)^3\\
\vdots \\
3(n-1)^2+3(n-1)+1 = ((n-1)+1)^3-(n-1)^3\\
3n^2+3n+1 = (n+1)^3-(n-1)^3\\
$$
The result of adding these formulas is
$$
3[1^2+2^2 + ... + (n-1)^2+n^2] + 3[1 + 2 + ... + (n-1)+n] + n = (n+1)^3-1^3
$$
I am able to follow up-to this point easily. I don't understand how this last equation goes from that to this $ n^3/3 + n^2/2 + n/6 = [1^2 + 2^2 + ... + n^2]$ I know thee sum of arithmetic series to n is $n(n-1)/2$ and that replaces the second expression on the left side.
Can some one please show me the algebra step by step?
 [/INST] $$ 3\left[1^2+2^2 + ... + (n-1)^2\right] + 3\left[1 + 2 + ... + (n-1)\right] + (n-1) = n^3-1^3 $$
$$3\left[1^2+2^2 + ... + (n-1)^2\right] + 3\left[1 + 2 + ... + (n-1)\right] + n = n^3 $$
$$3\left[1^2+2^2 + ... + (n-1)^2+n^2\right] + 3\left[1 + 2 + ... + (n-1)\right] + n = n^3+3n^2 $$
$$3\left[1^2+2^2 + ... + (n-1)^2+n^2\right] + 3\dfrac{(n-1)n}2 + n = n^3+3n^2 $$
$$3\left[1^2+2^2 + ... + (n-1)^2+n^2\right]  = n^3+3n^2-n-3\dfrac{(n-1)n}2 $$
$$1^2+2^2 + ... + (n-1)^2+n^2  = \dfrac{n^3}3+n^2-\dfrac n3-\dfrac{(n-1)n}2 $$
$$1^2+2^2 + ... + (n-1)^2+n^2  = \dfrac{n^3}3+\dfrac{n^2}2 +\dfrac{n}6 $$
</s>