Can I use this formula if the integers don't start at 1?

No, this formula is specific to integers starting at 1. If your integers start at a different number (like 5 to 20), you need the Arithmetic Series formula, or you can subtract the unwanted part: \frac{n(n+1)}{2} - \frac{(a-1)a}{2} where a is your starting point.

Why is there a division by 2 in the formula?

When you pair the smallest and largest terms (1 + n), the second-smallest and second-largest (2 + (n-1)), and so on, each pair totals n+1. There are n/2 such pairs, so the sum is \frac{n(n+1)}{2}. This pairing trick is often attributed to the mathematician Gauss, who discovered it as a student.

What if n is very large, like n = 1,000,000?

The formula works perfectly for any positive integer, no matter how large. It is actually the only practical way to find the sum since adding a million numbers would be impossible by hand. Just be careful with arithmetic: \frac{1000000 \cdot 1000001}{2} = 500000500000.

Is this the same as an arithmetic series?

It is a special case of the arithmetic series formula. An arithmetic series can have any starting point and any common difference, but this formula specifically handles the case where you start at 1 and increment by 1 each time. If you learn the general Arithmetic Series formula, you can derive this one as a special case by setting a_1 = 1 and d = 1.

Sum of First n Integers

Add all integers from 1 to n using the formula $\frac{n(n+1)}{2}$ instead of manually summing—much faster for large numbers.

\sum_{k=1}^{n} k = \frac{n(n+1)}{2}

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What each symbol means

What Sum of First n Integers takes

Symbol	Meaning
$n$	The positive integer that marks the end of your sequence; if $n$ is zero or negative, the formula does not apply.
$k$	A counting variable that represents each individual integer in the sequence from 1 to $n$ ; in the formula itself you don't substitute k directly, but it reminds you that you are summing all integer values from $k=1$ to $k=n$ .

When to use it

Reach for this formula whenever you need to find the sum of consecutive integers starting from 1.

Level

Usually taught in: Algebra II

Worked examples

1. Sum of integers 1 through 5

Problem

Find

1 + 2 + 3 + 4 + 5

$n = 5$
We identify that this is the sum of consecutive integers from 1 to 5, so $n = 5$ .
$\frac{n(n+1)}{2} = \frac{5(5+1)}{2}$
Substitute $n = 5$ into the formula.
$\frac{5 \cdot 6}{2} = \frac{30}{2} = 15$
Calculate $5 \times 6 = 30$ , then divide by 2 to get 15.

Answer: $15$

We recognise this as a sum of consecutive integers starting from 1, so we apply the formula directly with $n = 5$ . We can verify: adding manually gives $1 + 2 + 3 + 4 + 5 = 15$ , confirming our answer.

2. Sum of integers 1 through 50

Problem

Find the sum

1 + 2 + 3 + \cdots + 50

without adding each number individually.

$n = 50$
We have consecutive integers from 1 to 50, so $n = 50$ .
$\frac{n(n+1)}{2} = \frac{50(50+1)}{2}$
Substitute $n = 50$ into the formula.
$\frac{50 \cdot 51}{2} = \frac{2550}{2} = 1275$
Calculate $50 \times 51 = 2550$ , then divide by 2. Since one of $n$ or $n+1$ is always even, the division is always exact.

Answer: $1275$

For larger values of $n$ , the formula becomes essential—manually adding 50 numbers would be tedious and error-prone. The formula handles it in one step, and the key insight is that one of $n$ or $n+1$ is always even, so the product $n(n+1)$ is always divisible by 2.

3. Coffee shop daily sales over 30 days

Problem

A coffee shop tracks daily sales. On day 1 they earned $1, day 2 they earned $2, day 3 they earned $3, and so on. If this pattern continues for 30 days, what is the total revenue over the month?

$\text{Total} = 1 + 2 + 3 + \cdots + 30$
The coffee shop's earnings follow the pattern of consecutive integers from 1 to 30.
$n = 30$
Identify the endpoint of our sequence.
$\frac{30(30+1)}{2} = \frac{30 \cdot 31}{2}$
Apply the formula with $n = 30$ .
$\frac{930}{2} = 465$
Calculate $30 \times 31 = 930$ , then divide by 2.

Answer: $\$465$

The coffee shop's sales follow the pattern of consecutive integers, making this formula perfect for the problem. Computing the sum this way avoids having to add 30 separate daily figures, and in any scenario that generates consecutive integer patterns, this formula provides an instant answer.

Common mistakes

Where Sum of First n Integers usually goes wrong

The mistake	Why it is wrong	The fix
Writing $\sum_{k=1}^{n} k = n(n+1)$ and forgetting to divide by 2.	The division by 2 is essential: when you pair terms like $(1 + n)$ and $(2 + (n-1))$ , each pair sums to $n+1$ , and there are $n/2$ pairs, giving $\frac{n(n+1)}{2}$ .	Always include the division: the correct formula is $\frac{n(n+1)}{2}$ , never $n(n+1)$ .
Using the formula for a sequence that does not start at 1, such as $2 + 3 + 4 + \cdots + 10$ , and writing $\frac{10 \cdot 11}{2}$ .	This formula is specific to sequences starting at 1; applying it directly to a sequence starting at 2 gives the sum from 1 to 10 instead, including the 1 that was not part of the original problem.	For a sequence from $a$ to $n$ , compute $\frac{n(n+1)}{2} - \frac{(a-1)a}{2}$ , or use the Arithmetic Series formula instead.
Trying to use the formula when $n$ is zero, negative, or non-integer.	The formula is derived for positive integers; the reasoning behind it (pairing terms and counting pairs) breaks down outside this domain.	Always verify that $n$ is a positive whole number before applying the formula.

Tips and when to use something else

This formula is a special case of the Arithmetic Series formula $S = \frac{n(a_1 + a_n)}{2}$ , where $a_1 = 1$ and $a_n = n$ ; understanding the general formula helps you work with any arithmetic sequence.
For very large $n$ , compute $\frac{n}{2} \cdot (n+1)$ instead of $\frac{n(n+1)}{2}$ to avoid overflow in calculators—divide first, then multiply.
If the integers start at a number other than 1, use Arithmetic Series instead of trying to adapt this formula; it will save you from arithmetic mistakes.
Check your answer with a rough estimate: the sum is approximately $\frac{n^2}{2}$ , so for $n = 100$ you expect around 5000, and $\frac{100 \cdot 101}{2} = 5050$ matches perfectly.

Related formulas

Arithmetic SequenceFind the nth term in a sequence with constant differences between consecutive terms, or determine how many steps it takes to reach a specific value.Arithmetic SeriesFind the sum of an arithmetic sequence without adding every term—just use the first term, last term, and how many terms there are.Sigma NotationSigma notation compresses long sums into a compact formula, letting you write and calculate the sum of many terms efficiently and accurately.Recursive FormulaA recursive formula defines each term using the previous one, helping you find patterns and generate sequences step-by-step.

Frequently asked questions

Can I use this formula if the integers don't start at 1?: No, this formula is specific to integers starting at 1. If your integers start at a different number (like 5 to 20), you need the Arithmetic Series formula, or you can subtract the unwanted part: $\frac{n(n+1)}{2} - \frac{(a-1)a}{2}$ where $a$ is your starting point.
Why is there a division by 2 in the formula?: When you pair the smallest and largest terms ( $1 + n$ ), the second-smallest and second-largest ( $2 + (n-1)$ ), and so on, each pair totals $n+1$ . There are $n/2$ such pairs, so the sum is $\frac{n(n+1)}{2}$ . This pairing trick is often attributed to the mathematician Gauss, who discovered it as a student.
What if n is very large, like n = 1,000,000?: The formula works perfectly for any positive integer, no matter how large. It is actually the only practical way to find the sum since adding a million numbers would be impossible by hand. Just be careful with arithmetic: $\frac{1000000 \cdot 1000001}{2} = 500000500000$ .
Is this the same as an arithmetic series?: It is a special case of the arithmetic series formula. An arithmetic series can have any starting point and any common difference, but this formula specifically handles the case where you start at 1 and increment by 1 each time. If you learn the general Arithmetic Series formula, you can derive this one as a special case by setting $a_1 = 1$ and $d = 1$ .

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Reviewed 2026-09-18