Special Series

Sums of consecutive integers, consecutive squares, and consecutive cubes appear so often in contest problems that each has earned its own compact formula, saving the work of adding many terms by hand. These formulas also combine with subtraction to handle sums that do not start at 1, and one of them hides a striking identity connecting sums of cubes back to sums of integers.

What you will learn

• Recognize sums of consecutive integers, squares, and cubes
• Apply the standard formulas for these special sums
• Combine special series with other algebraic manipulation

Keep paper and a pencil nearby. Try each problem before opening a hint, and mark sections reviewed as you work through them.

The sum of consecutive integers, revisited

The pairing technique from the previous lesson gives the sum of the first $n$ positive integers as $1+2+3+\cdots+n=\frac{n(n+1)}{2}$. This is simply the arithmetic sum formula applied to the sequence $1,2,3,\ldots,n$, where the first term is 1, the last term is $n$, and there are $n$ terms.

This formula is worth memorizing on its own, since it is the building block that the sum-of-squares and sum-of-cubes formulas are built from and compared against.

The sum of consecutive squares

The sum of the first $n$ perfect squares has its own closed formula: $1^2+2^2+3^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6}$. Unlike the sum of integers, this formula is not derived by simple pairing, but it is used the same way — substitute the value of $n$ and simplify.

For example, the sum of the squares from 1 to 10 is $\frac{10\times11\times21}{6}=\frac{2310}{6}=385$. This formula grows much faster than the integer-sum formula, since it involves $n^3$ once the product is expanded, reflecting that squares themselves grow faster than the integers they come from.

The sum of consecutive cubes, and sums that don't start at 1

The sum of the first $n$ perfect cubes has a surprisingly elegant formula: $1^3+2^3+\cdots+n^3=\left(\frac{n(n+1)}{2}\right)^2$ — exactly the square of the sum-of-integers formula. This means the sum of the first $n$ cubes always equals $(1+2+\cdots+n)^2$, a striking identity that is worth remembering on its own.

Any of these three formulas can be adapted to a sum that does not start at 1 by subtracting: the sum from $a$ to $b$ equals the sum from 1 to $b$ minus the sum from 1 to $a-1$. For instance, $6^2+7^2+\cdots+12^2$ equals $(1^2+\cdots+12^2)-(1^2+\cdots+5^2)$, using the sum-of-squares formula for each piece and subtracting.

See the reasoning unfold

Pause after each step and predict what comes next.

Common mistakes to catch

• Using the sum-of-integers formula when a problem actually asks for a sum of squares or cubes.
• Forgetting to subtract the sum up to $a-1$, rather than up to $a$, when a sum starts at a number $a$ greater than 1.
• Confusing $\left(\frac{n(n+1)}{2}\right)^2$, the sum of cubes, with $\frac{n(n+1)(2n+1)}{6}$, the sum of squares.

Try with support

Guided practice

Choose a strategy, take a first step, and use the hints when you need direction.

Put it into practice

Independent practice

Work on your own first. If you reveal a solution, work through it and mark it reviewed.

Bring it together

• The sum of the first $n$ integers is $\frac{n(n+1)}{2}$, and the sum of the first $n$ squares is $\frac{n(n+1)(2n+1)}{6}$.
• The sum of the first $n$ cubes is $\left(\frac{n(n+1)}{2}\right)^2$, exactly the square of the sum of the first $n$ integers.
• A sum that starts above 1 is the difference of two full sums, computed with the same formula and subtracted.