Factorial, but with Addition

A factorial multiplies consecutive positive integers: for a positive integer $n$, $n! = 1\cdot 2\cdot 3\cdots n$. If you replace multiplication with addition, the usual interpretation is to add those same integers instead:

$$1+2+3+\cdots+n.$$

That sum is the nth triangular number, with formula $n(n+1)/2$. There is no universally standard notation called “addition factorial,” so it helps to write out the terms you mean. The formula below applies when the terms are the integers from 1 through a positive integer $n$.

Why the sum is $n(n+1)/2$

Let

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

Write the same sum in reverse order and add the two lines term by term:

$$\begin{aligned} S &= 1+2+3+\cdots+(n-1)+n,\\ S &= n+(n-1)+(n-2)+\cdots+2+1. \end{aligned}$$

Each pair of terms in a column adds to $n+1$, and there are $n$ columns. So $2S=n(n+1)$, which gives

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

The result is always an integer: one of two consecutive integers, $n$ or $n+1$, must be even.

Worked example: adding from 1 to 8

The sum is

$$1+2+3+4+5+6+7+8.$$

Using the formula with $n=8$ gives

$$\frac{8(8+1)}{2}=\frac{8\cdot 9}{2}=36.$$

You can check by pairing the ends: $1+8=9$, $2+7=9$, $3+6=9$, and $4+5=9$. There are four pairs, each totaling 9, so the sum is $4\cdot 9=36$.

For comparison, $8!$ is a product, not this sum:

$$8!=1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8=40{,}320.$$

If you meant a different sum

Two expressions that sound similar are often confused:

• Adding the integers through $n$: $1+2+\cdots+n=\frac{n(n+1)}{2}$.
• Adding the factorials through $n$: $1!+2!+\cdots+n!$.

The second expression is a sum of factorial values, not a factorial with addition replacing multiplication. For example, when $n=4$, it is $1!+2!+3!+4!=1+2+6+24=33$. The triangular-number formula does not apply to that sum.

You might also want to start at a number other than 1. For integers $a$ and $b$ with $a\le b$, the consecutive integers from $a$ through $b$, inclusive, number $b-a+1$. Their sum is

$$\frac{(a+b)(b-a+1)}{2}.$$

For instance, $4+5+6+7+8$ has five terms, so its sum is $(4+8)\cdot5/2=30$.

Common mistakes

• Using the factorial symbol for the sum: By convention, $n!$ means multiplication. Write the additions explicitly or call the result a triangular number.
• Leaving out an endpoint: The formula $n(n+1)/2$ includes both 1 and $n$. If the sum starts at 0, that extra term is zero, so the value is unchanged; if it starts at 2, subtract 1 from the formula.
• Using the wrong consecutive-integer formula: The sum from 1 to $n$ is $n(n+1)/2$, not $n(n-1)/2$. The latter is the sum from 1 to $n-1$.
• Applying it to factorial values: $1!+2!+\cdots+n!$ adds products, so it is a different calculation.

If your intended expression is “$n$ plus every integer below it down to 1,” the triangular-number formula gives the answer directly.