Finding the formula of sum $(k+1)^2+(k+2)^2+...(k+(n-1))^2$ I know the sum of square of numbers which stars from $1$
but I don't know what the formula becomes when the first term is not $1$ as follow
$$(k+1)^2+(k+2)^2+...(k+(n-1))^2$$

Hint
$$\sum_{i = k}^n i^2 = \sum_{i=1}^n i^2 - \sum_{i=1}^{k-1}i^2$$
$\text{Now substitute using the fact that }$
$$\sum_{i=1}^n i^2 = \frac{n(n + 1)(2n + 1)}{6}$$
