1 + 2 + 3 + ... + n = n(n+1)/2: Gauss's formula

The formula

$1 + 2 + 3 + \ldots + n = \sum_{k=1}^{n} k = \dfrac{n(n+1)}{2}$

Young Gauss’s trick

Write the sum in order, then in reverse:

$S = 1 + 2 + 3 + \ldots + n$ $S = n + (n-1) + (n-2) + \ldots + 1$

Add the two lines term by term:

$2S = (n+1) + (n+1) + \ldots + (n+1) = n(n+1)$

So $S = \dfrac{n(n+1)}{2}$.

Quick applications

1 + 2 + … + 100 = $\dfrac{100 \times 101}{2} = 5050$.

Sum of integers from $p$ to $n$ (with $p \leq n$): $p + (p+1) + \ldots + n = \dfrac{(n+p)(n-p+1)}{2}$

Number of terms × average: $S = (\text{\# terms}) \times \dfrac{\text{first} + \text{last}}{2}$

Related formulas to know

Sum of squares: $\sum_{k=1}^{n} k^2 = \dfrac{n(n+1)(2n+1)}{6}$

Sum of cubes (beautiful!): $\sum_{k=1}^{n} k^3 = \left( \dfrac{n(n+1)}{2} \right)^2 = \left( \sum_{k=1}^{n} k \right)^2$

Induction proof

Base: $n=1$, $S_1 = 1 = 1 \cdot 2 / 2$. ✓

Step: if $S_n = n(n+1)/2$, then $S_{n+1} = S_n + (n+1) = \dfrac{n(n+1)}{2} + (n+1) = \dfrac{(n+1)(n+2)}{2}$ ✓

Common mistakes

1. Dividing by 3 instead of 2.
2. Using $(n+1)/2$ and forgetting the $n$.
3. Confusing number of terms and $n$: from $1$ to $n$ = $n$ terms; from $p$ to $n$ = $n - p + 1$ terms.

?À toi de jouer

1 + 2 + 3 + ... + 50 = ?

A.1275B.2550C.50D.1250

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