Summation $\sum_{j=2}^{n-1}j^2$ Properties

Question

I'm dealing with something like $\sum_{j=2}^{n-1}j^2$. I know I can do this $\sum_{j=1}^{n-1}j^2 - \sum_{j=1}^{1}j^2$.

Would that be equal to $\frac{j(j+1)(2j+1)}{6} - j^2$ or I'm missing some properties with $n-1$?

If so, which ones?

Answer 1

Since

$$\sum_{r=1}^mr^2=\frac{m(m+1)(2m+1)}{6},$$

by letting $\color{blue}{m=n-1}$, we have,

$$\begin{align} \sum_{j=2}^{n-1}j^2&=\sum_{j=1}^{n-1}j^2-\color{green}{\sum_{j=1}^1j^2}\\ &=\frac{\color{blue}{(n-1)}(\color{blue}{(n-1)}+1)(2\color{blue}{(n-1)}+1)}{6}-\color{green}{1^2}\\ &=\frac{(n-1)n(2n-1)}{6}-\color{green}{1}. \end{align}$$

Answer 2

$$\sum_{j=2}^{n-1}j^2=\sum_{j=1}^{n}j^2-1^2-n^2=\frac{n(n+1)(2n+1)}{6}-1-n^2$$