Where the Consecutive Sum Formula Comes From

Where the Consecutive Sum Formula Comes From — Moosa Academy

Most students meet $1 + 2 + 3 + \cdots + n = \dfrac{n(n+1)}{2}$ as a rule to memorise. It is worth seeing where it comes from, because the argument is short, entirely visual, and explains why the $2$ sits underneath.

Concept Pair the first with the last

To add $1$ through $10$ , do not work left to right. Pair the smallest with the largest, then work inward:

1 + 10 = 11

2 + 9 = 11

3 + 8 = 11

4 + 7 = 11

5 + 6 = 11

Every pair totals the same $11$ , and there are $5$ of them:

5 \times 11 = 55

The pairs match because as one number climbs by $1$ , its partner falls by $1$ . Their total cannot change.

Example What if one is left over?

Add $1$ through $11$ . There is now an odd count of numbers, so the middle one has no partner.

Pairs:

1 + 11

2 + 10

3 + 9

4 + 8

5 + 7

— five pairs of

12

5 \times 12 = 60

The middle number

6

is left over

60 + 6 = 66

The sum is $66$

The method still works, but it now has two cases and a leftover to remember. The next idea removes that awkwardness completely.

Theorem Write the sum twice

Writing the sum forwards and backwards, then adding the two lines column by column, gives $1 + 2 + \cdots + n = \dfrac{n(n+1)}{2}$ .

Call the sum $S$ and write it out twice, the second time in reverse:

S = 1 + 2 + 3 + \cdots + n

S = n + \cdots + 3 + 2 + 1

Add the two lines. Each column pairs a number with its mirror, and every column totals $n + 1$ :

2S = (n+1) + (n+1) + \cdots + (n+1)

There are

n

columns, so

2S = n(n+1)

Divide both sides by

2

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

Doubling the sum is what makes the leftover problem vanish. With two copies there is always an even number of terms, so nothing is ever stranded — and the $2$ in the denominator is simply undoing that doubling.

Example Checking against both cases

A derivation is only convincing if it reproduces the answers found by hand.

n = 10

\dfrac{10 \times 11}{2} = \dfrac{110}{2} = 55

— matches the pairing result

n = 11

\dfrac{11 \times 12}{2} = \dfrac{132}{2} = 66

— matches the leftover result

Both cases agree, even and odd alike

Notice that one of $n$ and $n+1$ is always even, so the division by $2$ never leaves a fraction. The formula always returns a whole number, as a count of this kind must.

Summary

Pairing the first term with the last gives pairs of equal total.
Pairs match because one number rises exactly as the other falls.
With an odd count of numbers the middle term is left unpaired and must be added separately.
Writing the sum forwards and backwards avoids that problem entirely.
Every column then totals $n+1$ , and there are $n$ columns, so $2S = n(n+1)$ .
Halving gives $S = \dfrac{n(n+1)}{2}$ , which works for every $n$ .