Sums of integer powers

For a positive integer n, the sums from r = 1 to n are Σr = n(n + 1)/2 and Σr² = n(n + 1)(2n + 1)/6. A constant summand c contributes nc because there are n terms. Expand a polynomial summand before combining these formulas, keeping the same lower and upper limits on each sum.

When the first index is greater than one, subtract the unwanted initial terms. For example a sum from r = 4 to n equals the sum from 1 to n minus the sum from 1 to 3. Check a proposed formula at a small n to catch an indexing error, then use a proof such as induction when a general justification is requested.

Worked example

Find Σ(2r² − 3r + 1) for r = 1,…,5.

1. The square sum is 5 × 6 × 11/6 = 55, and the linear sum is 5 × 6/2 = 15.
2. There are five constant terms, so their sum is 5.
3. Combine: 2 × 55 − 3 × 15 + 5.

Answer: 70.

Revise first: Expanding and factorising.