Skip to content

Sums of odd numbers

Patterns and sequences · level 4, lesson 6 of 6

A question to start

What is \(1 + 3 + 5 + 7 + \cdots + 99\), all the odd numbers up to 99?

That is fifty numbers to add. With a calculator it takes a minute and one slip spoils it. In level 3 you met the pattern that sums of odd numbers starting from 1 are square numbers, and you saw it for small squares. But level 3 also warned you that patterns can trick you. In this lesson you prove the pattern for every size at once, three different ways, and then the answer, 2500, takes five seconds.

The idea

The \(n\)th odd number

The odd numbers are 1, 3, 5, 7, 9, ... They go up by 2 each time, so by lesson 2 their position rule has \(2n\) in it. The zero term is \(1 - 2 = -1\), so

\[\text{the } n\text{th odd number} = 2n - 1\]

There is a simpler way to see it: the \(n\)th even number is \(2n\) (2, 4, 6, ...), and each odd number is one less than the even number in the same position. Check: the 5th odd number is \(2 \times 5 - 1 = 9\). ✓ Running it backwards: 99 is the \(n\)th odd number when \(2n - 1 = 99\), so \(n = 50\). There are 50 odd numbers from 1 to 99.

The claim

Adding the first \(n\) odd numbers gives \(n\) squared:

\[1 + 3 + 5 + \cdots + (2n - 1) = n^2\]

The "\(\cdots\)" means "and all the odd numbers in between". In level 3 we checked this for small \(n\): \(1 = 1\), \(1 + 3 = 4\), \(1 + 3 + 5 = 9\), \(1 + 3 + 5 + 7 = 16\). Checking is not proving. Here are three proofs, each of which works for every \(n\).

Proof 1: L-shapes (a picture proof)

1 3 5 7 9 1 + 3 + 5 + 7 + 9 each L wraps the square before 5 L-shapes make a 5 by 5 square so the sum is 5 × 5 = 25

Build an \(n\) by \(n\) square out of L-shaped layers, starting from the corner. The L that turns a \((k - 1)\) by \((k - 1)\) square into a \(k\) by \(k\) square has a bottom row of \(k\) squares and a side column of \(k - 1\) more (the corner square is already in the row). That is \(k + (k - 1) = 2k - 1\) squares: the \(k\)th odd number.

So the 1st L has 1 square, the 2nd has 3, the 3rd has 5, ..., and the \(n\)th has \(2n - 1\). Together the \(n\) L's fill the \(n\) by \(n\) square exactly, with no gaps and no overlaps. Therefore the first \(n\) odd numbers add to \(n^2\).

Why is this a proof and not just a check? Because nothing in the argument depended on \(n\) being 5, or 10. The description of the \(k\)th L works for any \(k\), and the stacking works for any \(n\). The picture shows one size, but the reasoning is about all sizes.

Proof 2: pairing the ends

Write the sum and pair the first number with the last, the second with the second-last, and so on:

  • \(1 + (2n - 1) = 2n\)
  • \(3 + (2n - 3) = 2n\)
  • \(5 + (2n - 5) = 2n\)

Every pair adds to \(2n\): as you move in from the ends, one number goes up by 2 and the other goes down by 2, so the pair total stays the same. A clean way to count the pairs is to write the sum twice, once forwards and once backwards, and add them term by term:

\[ \begin{aligned} S &= 1 + 3 + \cdots + (2n - 3) + (2n - 1)\\ S &= (2n - 1) + (2n - 3) + \cdots + 3 + 1\\ 2S &= 2n + 2n + \cdots + 2n + 2n \end{aligned} \]

There are \(n\) columns, each adding to \(2n\), so \(2S = n \times 2n = 2n^2\) and \(S = n^2\). This works whether \(n\) is even or odd: writing the sum twice avoids any worry about a middle number with no partner.

Proof 3: a pyramid, made level

rows 1, 3, 5, 7: 16 squares pair from the ends: 1 + 7 = 8 and 3 + 5 = 8 2 pairs of 8 = 16 = 4 × 4

Stack rows of 1, 3, 5 and 7 squares, centred, to make a pyramid. Pair the top row with the bottom row (\(1 + 7 = 8\)) and the second row with the third (\(3 + 5 = 8\)). Each pair makes a row of 8, which is \(2n\), and there are \(n \div 2\) pairs. So the 16 squares rearrange into 2 rows of 8, and \(2 \times 8 = 16 = 4 \times 4\). (This is proof 2 again, but seen as a shape. When \(n\) is odd, the middle row has \(n\) squares on its own, half of \(2n\), and the count still works.)

Using the result

Sums that do not start at 1. To add a run of odd numbers in the middle, like \(11 + 13 + \cdots + 19\), take the sum from 1 up to 19, and remove the sum from 1 up to 9:

  • 19 is the 10th odd number, so \(1 + \cdots + 19 = 10^2 = 100\).
  • 9 is the 5th odd number, so \(1 + \cdots + 9 = 5^2 = 25\).
  • So \(11 + 13 + 15 + 17 + 19 = 100 - 25 = 75\).

Sums of even numbers. Each even number is 1 more than an odd number: \(2 = 1 + 1\), \(4 = 3 + 1\), and so on. So the first \(n\) even numbers add to the first \(n\) odd numbers plus \(n\) extra ones:

\[2 + 4 + \cdots + 2n = n^2 + n = n(n + 1)\]

That is twice the \(n\)th triangular number from level 3, which makes sense: \(2 + 4 + \cdots + 2n\) is \(2 \times (1 + 2 + \cdots + n)\).

Worked examples

1. The opening question. 99 is the 50th odd number, so \(1 + 3 + \cdots + 99 = 50^2 = 2500\).

2. The first 12 odd numbers. \(12^2 = 144\). (The last one is \(2 \times 12 - 1 = 23\).)

3. \(21 + 23 + \cdots + 39\).

  • 39 is the 20th odd number: \(1 + \cdots + 39 = 400\).
  • 19 is the 10th odd number: \(1 + \cdots + 19 = 100\).
  • So the sum is \(400 - 100 = 300\). (There are 10 numbers, and their average is \((21 + 39) \div 2 = 30\), so \(10 \times 30 = 300\) ✓.)

4. The even numbers up to 100. \(2 + 4 + \cdots + 100\) has 50 terms: \(50 \times 51 = 2550\).

5. Running it backwards. You add odd numbers, starting at 1, until the total is exactly 400. How many did you add, and what was the last one? \(n^2 = 400\) gives \(n = 20\), and the last is \(2 \times 20 - 1 = 39\).

Try it

  1. Work out \(1 + 3 + 5 + \cdots + 19\).
  2. What is the 15th odd number?
  3. Work out the sum of the first 30 odd numbers.
  4. Work out \(1 + 3 + 5 + \cdots + 79\).
  5. Work out \(11 + 13 + \cdots + 29\).
  6. Work out \(2 + 4 + 6 + \cdots + 40\).
  7. Write the odd numbers in rows: 1 | 3 + 5 | 7 + 9 + 11 | 13 + 15 + 17 + 19 | ... Add up each row. What do you notice? Predict the next row's total and check it.
  8. Use the L-shape picture to explain why \(n^2 - (n - 1)^2 = 2n - 1\). Then use it to find \(50^2 - 49^2\) without squaring.
Answers
  1. 19 is the 10th odd number (\(2 \times 10 - 1 = 19\)), so the sum is \(10^2 = 100\).
  2. \(2 \times 15 - 1 = 29\).
  3. \(30^2 = 900\).
  4. \(2n - 1 = 79\) gives \(n = 40\): \(40^2 = 1600\).
  5. 29 is the 15th odd number and 9 is the 5th: \(15^2 - 5^2 = 225 - 25 = 200\).
  6. 20 terms: \(20 \times 21 = 420\).
  7. The row totals are 1, 8, 27, 64: these are \(1 \times 1 \times 1\), \(2 \times 2 \times 2\), \(3 \times 3 \times 3\), \(4 \times 4 \times 4\) (the cube numbers). The next row is \(21 + 23 + 25 + 27 + 29 = 125 = 5 \times 5 \times 5\). ✓ (Proving this for every row is a good challenge; level 7 of this topic uses it.)
  8. An \(n\) by \(n\) square is an \((n - 1)\) by \((n - 1)\) square plus one more L-shape, the \(n\)th, which has \(2n - 1\) squares. So the difference between the two squares is \(2n - 1\). For \(n = 50\): \(50^2 - 49^2 = 99\). (Check: \(2500 - 2401 = 99\) ✓.)

Common mistake

Squaring the last number instead of the count. \(1 + 3 + \cdots + 19\) is not \(19^2\). The rule squares the number of terms, \(n\), and 19 is only the 10th odd number. First find \(n\) from \(2n - 1 = \text{last}\), then square \(n\).

Calling a few checks a proof. "It works for 1, 2, 3, 4 and 5" is evidence, not proof. The L-shape argument is a proof because it explains why the \(k\)th layer always has \(2k - 1\) squares, for any \(k\).

Where it leads