Sum of the first \(n\) integers
\(1 + 2 + 3 + ... + n = \)
$$\sum_{k=1}^n k = \frac{1}{2}n(n+1)$$
Use cases: Finding a missing number from \(1 ... n\)
Sum of the squares of the first \(n\) integers
\(1^2 + 2^2 + 3^2 + ... + n^2 = \)
$$\sum_{k=1}^n k^2 = \frac{1}{6}n(n+1)(2n+1)$$
Growth of functions
