AlgebraAMC 10

Sums of Powers

\sum k = \tfrac{n(n+1)}{2},\quad \sum k^2 = \tfrac{n(n+1)(2n+1)}{6},\quad \sum k^3 = \left(\tfrac{n(n+1)}{2}\right)^2

Sums from $k = 1$ to $n$ .

What it says

These give closed forms for adding the first $n$ integers, squares, or cubes. Notice that the sum of the first $n$ cubes is exactly the square of the sum of the first $n$ integers. Counting squares in a grid is a classic place where the sum of squares shows up.

Show proof

The first is the arithmetic series formula. For squares, sum the identity $(k+1)^3 - k^3 = 3k^2 + 3k + 1$ from $k = 1$ to $n$ : the left side telescopes to $(n+1)^3 - 1$ , then solve for $\sum k^2$ . The cube formula follows by induction: adding $(n+1)^3$ to $\left(\frac{n(n+1)}{2}\right)^2$ gives $\left(\frac{(n+1)(n+2)}{2}\right)^2$ .

When to use it

Adding consecutive integers, squares, or cubes, including counting squares on a grid.

Example

Evaluate $1^2 + 2^2 + 3^2 + \cdots + 10^2$ .

Show solution

$\frac{10\cdot11\cdot21}{6} = 385$ . Answer: $385$