The sum of the first n terms

A story about a quick answer

About two hundred years ago, a teacher in Germany wanted some quiet. He told his class to add all the whole numbers from $1$ to $100$. Within a minute, a boy called Carl Friedrich Gauss put his slate on the desk with the answer: $5050$.

Gauss did not add the numbers one by one. He paired them:

There are $50$ such pairs, each adding to $101$. So the sum is $50 \times 101 = 5050$.

The numbers $1, 2, 3, \ldots, 100$ form an arithmetic progression. Gauss's trick works for every arithmetic progression.

Pairing in a picture

The blue staircase is $1 + 2 + 3 + 4 + 5 + 6$. Turn a copy upside down and put it on top. Every column is now $7$ squares tall, and there are $6$ columns. The two staircases together hold $6 \times 7 = 42$ squares, so one staircase holds $21$.

The formula

Let the arithmetic progression have first term $a$, common difference $d$ and $n$ terms, with last term $l$. Write the sum twice, once forwards and once backwards:

Add the two lines column by column. Each column adds to $a + l$, and there are $n$ columns:

The last term is $l = a + (n - 1)d$. Put that in, and we get the formula to use when $l$ is not given:

Using the formula

Find the sum of the first $15$ terms of $7, 11, 15, 19, \ldots$

1. Read the numbers. $a = 7$, $d = 11 - 7 = 4$, $n = 15$.
2. Use the formula. $S_{15} = \dfrac{15}{2}[2 \times 7 + 14 \times 4] = \dfrac{15}{2}[14 + 56] = \dfrac{15}{2} \times 70$.
3. Work it out. $S_{15} = 15 \times 35 = 525$.

Check with the other form. The last term is $l = 7 + 14 \times 4 = 63$, and $\dfrac{15}{2}(7 + 63) = \dfrac{15}{2} \times 70 = 525$. Both forms agree.

The first n natural numbers

For $1, 2, 3, \ldots, n$ we have $a = 1$ and $l = n$. So

With $n = 100$ this gives $\dfrac{100 \times 101}{2} = 5050$, Gauss's answer.

Practise with new numbers

Try it in Python

The program adds the same arithmetic progression in three ways: one term at a time, by Gauss's pairs, and by the formula.

Common mistakes

• Leaving out the half. The formula is $\dfrac{n}{2}[2a + (n - 1)d]$. Without the $\tfrac{1}{2}$, the answer is twice too big.
• Using $n$ in place of $n - 1$. The last term is $a + (n - 1)d$, because the first term has no $d$ added.
• Mixing up $n$ and $l$. $n$ counts the terms, and $l$ is the value of the last term.
• Losing a negative $d$. For $40, 37, 34, \ldots$, $d = -3$, and $(n - 1)d$ is negative.

Remember these

• $S_n = \dfrac{n}{2}(a + l)$, when the last term $l$ is known.
• $S_n = \dfrac{n}{2}[2a + (n - 1)d]$, when $a$ and $d$ are known.
• $1 + 2 + 3 + \cdots + n = \dfrac{n(n + 1)}{2}$.
• Check a sum with the other form of the formula.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Write the sum of the first $n$ terms when you know $a$, $n$ and the last term $l$.

2. Find the sum of the first $10$ terms of $6, 9, 12, \ldots$

3. What is $1 + 2 + 3 + \cdots + 60$?

4. Ajay adds $2 + 4 + 6 + \cdots + 40$ and says there are $40$ terms. Is he right?

5. Why does pairing the first term with the last always give the same total as the second term with the second last?