When Carl was 10, his arithmetic teacher, wishing to keep the class occupied, set the pupils the task of finding

1 + 2 + 3 + ... + 100.

Almost immediately Carl wrote down an answer and laid his slate on his desk.

The other pupils toiled on for the rest of the hour. At the end of the period it was found that Carl alone had the direct answer, and he was able to explain how he had arrived at the result. His explanation was very simple. How did he do it?

(The above story, with variations, is told about the brilliant mathematician Carl Friedrich Gauss who lived 1777–1855).

Hint 1

Perhaps the given sum can be written in a different order, and the two sums combined in a simple way?

Hint 2

Or equivalently, perhaps pairs of numbers in the given sum can be added together to the same total?

Solution

The two hints are in fact equivalent. Suppose we denote the required sum by S.

Then
        S = 1 + 2 + 3 + ... + 98 + 99 + 100.
As well,
        S = 100 + 99 + 98 + ... + 3 + 2 + 1.

If we add the corresponding terms of the two sums together, we obtain
     2S = (1 + 100) + (2 + 99) + ... + (100 + 1)
          = 100 x 101,

since each bracketed pair adds to 101 and there are 100 pairs.

Thus     S =  ^{100 x 101} / ₂  =  5050.

The alternative method is to note that

    S = (1 + 100) + (2 + 99) + ... + (50 + 51)
        = 50 x 101   =   5050 as before.

Extensions

1. Can you use the same method to show that

          1 + 2 + 3 + ... + n = ^{n(n +1)}/₂?

2. You are Carl. Your teacher asks you to add up

          101 + 102 + 103 + ... + 200

in a hurry. Can you do it?

3. You are asked to find the sum of
          ¹/₂ + ¹/₆ +¹/₁₂ + ¹/₂₀ + ¹/₃₀ + ... + ¹/₁₀₀₁₀ .
Look at the factors of each denominator, and hence split each fraction into the difference of two fractions. The required sum follows immediately.

Hint 1

Hint 2

Solution

Extensions