The sum of the first $n$ squares

Add up the square numbers:

1^2 + 2^2 + 3^2 + \dots + n^2 = \frac{n(n + 1)(2n + 1)}{6}

Check it with $n = 4$ . The left side is $1 + 4 + 9 + 16 = 30$ , and the right side is $\dfrac{4 \times 5 \times 9}{6} = 30$ . It works.

This page proves it for every $n$ at once. The method is proof by induction, which is how a statement about all the counting numbers is usually proved.

The shape of an induction proof

Induction has two steps, and you need both:

Show it works for the first case. Here, $n = 1$ .
Show that if it works for one case, it must work for the next. That is: if the formula is true for some $n$ , then it is true for $n + 1$ .

Together those cover everything. Step 1 gets you to $n = 1$ . Step 2 then carries you from 1 to 2, from 2 to 3, from 3 to 4, and onward without end. It is worth seeing why one step without the other proves nothing: a formula that is true only for $n = 1$ passes step 1 and fails step 2, and a formula that is always one step behind passes step 2 and fails step 1.

Step 1: the first case

We need

1^2 = \frac{1 \times 2 \times 3}{6}

The right side is $\dfrac{6}{6} = 1$ . So the statement is true when $n = 1$ .

Step 2: the inductive step

Now suppose the formula is true for some particular $n$ . That is our assumption:

1^2 + 2^2 + \dots + n^2 = \frac{n(n + 1)(2n + 1)}{6}

We have to show the formula holds for $n + 1$ , which means showing

1^2 + 2^2 + \dots + n^2 + (n + 1)^2 = \frac{(n + 1)(n + 2)(2n + 3)}{6}

Start with the left side and use the assumption to replace the long sum:

1^2 + \dots + n^2 + (n + 1)^2 = \frac{n(n + 1)(2n + 1)}{6} + (n + 1)^2

Now take out the common factor $(n + 1)$ from both terms:

= (n + 1)\left[\frac{n(2n + 1)}{6} + (n + 1)\right]

Put the second term over $6$ so the bracket is a single fraction:

= (n + 1)\left[\frac{n(2n + 1) + 6(n + 1)}{6}\right]

Expand the top:

= (n + 1)\left[\frac{2n^2 + n + 6n + 6}{6}\right] = (n + 1)\left[\frac{2n^2 + 7n + 6}{6}\right]

The quadratic on top factorises, because $2n^2 + 7n + 6 = (2n + 3)(n + 2)$ :

= \frac{(n + 1)(n + 2)(2n + 3)}{6}

That is exactly the formula with $n + 1$ in place of $n$ , because $2(n + 1) + 1 = 2n + 3$ .

Both steps are done. The statement is true for $n = 1$ , and true for $n+1$ whenever it is true for $n$ . So it is true for every counting number.

What each step is doing, and what it is not

The inductive step does not prove the formula for any particular number. It proves a link between one case and the next. On its own it would be satisfied by a formula that is wrong everywhere, as long as the wrongness matched up.

That is why step 1 is not a formality. It is the anchor: it is the one place the formula is checked against reality, and the link in step 2 carries that check onward through every number.

A common error

Watch for this one, because it is easy to do in an exam and it costs the mark.

When you write "suppose the statement is true for $n$ ", you must then use that supposition. A proof that subtracts the two sides and gets zero has usually assumed the very thing it was asked to prove.

The safe structure is the one above: start from the left side, put the assumption in, and arrive at the right side of the $n+1$ statement. The direction of the work is what makes it a proof rather than a check.

Check your understanding

Use the formula to work out $1^2 + 2^2 + \dots + 10^2$ .
In step 1, why is it enough to check $n = 1$ rather than $n = 0$ ?
Show that if $\dfrac{n(n+1)(2n+1)}{6}$ is a whole number, then so is $\dfrac{(n+1)(n+2)(2n+3)}{6}$ . What does that tell you about every value of this sum?

Answers

$\dfrac{10 \times 11 \times 21}{6} = \dfrac{2310}{6} = 385$ .
The statement is about $1^2 + 2^2 + \dots + n^2$ , which begins at $1$ and has no $0^2$ term, so $n = 1$ is its first case. Starting at $0$ would prove something slightly different.
From step 2 the two fractions are equal, so the second is a whole number exactly when the first is. Starting from $n = 1$ , which is a whole number, the step carries that along: the sum of the first $n$ squares is always a whole number.

The sum of the first nn squares