Zero to Theorem: The sum of an arithmetic series

This theorem is one of the oldest mathematical propositions, dating back to at least 1700 BCE. It establishes an algorithm for calculating sums of numbers which differ only by multiples of a fixed value. See if you can construct a similar algorithm.

  1. addition: the act of combining two or more expressions
  2. sum: the result of adding two or more expressions
  3. sequence: a list of expressions
  4. series: the mapping of a sequence to its sum
  5. arithmetic series: a series whose terms differ only by multiples of a fixed value
  6. Find the sum of 1 + 2 + 3. Analysis
  7. See if the last observation applies to 1 + 2 + 3 + 4. Analysis
  8. Find the sum of 2 + 4 + 6 + ⋯ + 2n. Solution
  9. Find the sum of 1 + 3 + 5 + ... + 2n - 1.
  10. Find the sum of 1 + 2 + 3 + ... + n.
  11. Find the sum of 1 + 4 + 7 + ... + 3n - 2.
  12. Find the sum of 1 + 5 + 9 + ... + 4n - 3.
  13. Find the sum of 5 + 9 + ... + 4n - 3.
  14. Find the sum of 1 + (1 + d) + (1 + 2d) + ... + (1 + nd).
  15. Find the sum of a + (a + d) + (a + 2d) + ... + (a + nd).

1 + 2 + 3 = 6, yes, but there is more here than meets the eye. Some observations: (1) the sum is three times the middle number; 1 + 2 + 3 = 3 ⋅ 2 and (2) the sum is three times the average of the first and last numbers; 1 + 2 + 3 = 3(1 + 3) / 2. Will this work on other arithmetic sequences?

2 + 4 + 6 + ... + 2n can be understood by taking partial sums. Let Ek be the partial sum of the first k even numbers. We see that E1 = 2, E2 = 6, E3 = 12, etc. With a little analysis, we notice that k is a factor of Ek in the first three cases, so it seems that En = n(n + 1). To prove this, we must use the principle of mathematical induction. Assuming that it works for an arbitrary integer n, we check that it works for n + 1. 2 + 4 + 6 + ... + 2(n + 1) = 2 + 4 + 6 + ... + 2n + 2(n + 1). By substitution, we get 2 + 4 + 6 + ... + 2(n + 1) = n(n + 1) + 2(n + 1). Since n(n + 1) + 2(n + 1) = (n + 1)(n + 2), we see 2 + 4 + 6 + ... + 2(n + 1) = (n + 1)(n + 2). Therefore, it works for all integers.

Let On represent the sum of the first n odd numbers. Observe that 1 + 3 + 5 + ... + 2n - 1 = (2 - 1) + (4 - 1) + (6 - 1) + ... + (2n - 1). Thus, we see that On = En - n. Since En = n(n + 1), we get On = n(n + 1) - n by substitution. Therefore, On = n2.

Let Tn = 1 + 2 + 3 + ... + n. For this, we will employ Gauss' ingenius argument. Observe that 2Tn = En. Therefore, Tn = n(n + 1) / 2.

By substitution, we have

a + (a + d) + (a + 2d) + ... + (a + nd) = (n + 1)a + dTn = na + a + dn(n + 1) / 2 = (n + 1)(2a + dn) / 2. Since an = a + (n + 1)d