Add all the numbers up to 100

Let's prove the general form of this:

Let us have the sum of the first $n$ positive integers, i.e. $S_n = 1+2+3+\cdots+n$ .
Now clearly $2S_n = (1 + n) + (2 + n - 1) + (3 + n - 2) + \cdots + (n + 1) = n(n+1) \implies S_n = \dfrac{n(n+1)}2. \square$

Substituting $n = 100$ gives $1 + 2 + 3 + \cdots + 100 = \dfrac{100\cdot101}2 = 50 \cdot 101 = \boxed{5050}$ , which is the answer.

Relevant wiki: Sum of n, n², or n³

Pairing numbers is a common approach to this problem. Instead of writing all the numbers in a single column, let’s wrap the numbers around, like this: 1 2 3 4 5 10 9 8 7 6 An interesting pattern emerges: the sum of each column is 11. As the top row increases, the bottom row decreases, so the sum stays the same.

Because 1 is paired with 10 (our n), we can say that each column has (n+1). And how many pairs do we have? Well, we have 2 equal rows, we must have n/2 pairs. Number of pairs * Sum of each pair = (n/2)(n+1) = n(n+1)/2 which is the formula.

Relevant wiki: Arithmetic Progression Sum

$S = 1 + 2 + 3 + \cdots + 100 \\ {\text{Given series is an arithmetic progression where, }} a = 1, d = 1, n = 100,l = 100. \\ S = \dfrac{n}{2}(a + l) = \dfrac{100}{2}(100 + 1) = 50 \times 101 = \color{#E81990}{\boxed{5050}}$

$\frac{100^2*100}{2}=5050$