@rkenmi - Math Formulas for C.S.

# 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

$$1^2 + 2^2 + 3^2 + ... + n^2 =$$

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