A Formula for the Sum of the First n Natural Numbers

# A Formula for the Sum of the First n Natural Numbers

Theorem 1: For every natural number $n$, $\displaystyle{1 + 2 + ... + n = \frac{n(n+1)}{2}}$. |

**Proof (Pictoral):**

- Represent each natural number $n$ by $n$-many unit squares. Then each natural number is equal to the area of its unit squares. For each $n$ we can construct a "pyramid" that is $n$ squares high in the manner as depicted above. Then the area of the $n^{\mathrm{th}}$ pyramid is the sum $1 + 2 + ... + n$.

- But the $n^{\mathrm{th}}$ pyramid is contained in the $n(n+1)$ rectangle and clearly its area is half the rectangle, so:

\begin{align} \quad 1 + 2 + ... + n = \frac{n(n+1)}{2} \quad \blacksquare \end{align}

**Proof (Algebraic):**We prove this by mathematical induction. Let $P(n)$ be the statement that the sum of the first $n$ natural numbers is equal to $\displaystyle{\frac{n(n+1)}{2}}$. If $n = 1$ then $P(1)$ says that the sum of the first $1$ natural numbers is equal to $\displaystyle{\frac{1(2)}{2} = 1}$ which is true.

- Assume that for some $k$ that $P(k)$ is true. We want to show that $P(k + 1)$ is true. We have that the sum of the first $k+1$ number is:

\begin{align} \quad 1 + 2 + ... + k + (k+1) &= [1 + 2 + ... + k] + (k + 1) \\ \end{align}

- By assumption $P(k)$ is true and so $\displaystyle{1 + 2 + ... + k = \frac{k(k+1)}{2}}$. So:

\begin{align} \quad 1 + 2 + ... + k + (k+1) &= \frac{k(k+1)}{2} + (k+1) \\ &= \frac{k^2 + k}{2} + (k + 1) \\ &= \frac{1}{2}k^2 + \frac{1}{2}k + k + 1 \\ &= \frac{k^2 + 3k + 2}{2} \\ &= \frac{(k + 1)([k + 1] + 1)}{2} \end{align}

- So $P(k+1)$ is true. By the principle of mathematical induction we have that $P(n)$ is true for every natural number $n$. $\blacksquare$