Example 1. What is the sum 1 + 2 + 3 +···+ 99 + 100, of all the positive integers from 1 to 100?
Example 2. Let f : R → R be a function satisfies the equation f(x) + f(y) = f(x + y) for all x, y ∈ R. Prove that f is an odd function.
Of course, you can find the solution just below, but it is highly recommended that, you first try to solve it on your own.
Just remember the words of Paul Halmos, who says “the only way to learn mathematics is to do mathematics”.
Solution 1. Basically, it is the sum of first 100 natural numbers. Now, pair up the numbers as below:
(100 + 1) = (99 + 2) = (98 + 3) = ··· = (51 + 50) = 101 [one number from the beginning and one from the end]
One can check that, there are 50 such pairs and when we add the numbers in each pairs, we get a sum 101.
Evidently, the sum is 101 × 50 = 5050.
We can represent it in the following way also. Assume the given sum as S. Then
S = 1 + 2 + 3 +···+ 99 + 100
S = 100 + 99 + 98 +···+ 2 + 1
[In the second line, just rearrange the given sum and write it as above]
Then on addition, we get
2S = 101 + 101 + 101 + ··· + 101 + 101 = (100)(101).
Required sum = 101 × 50 = 5050, as before.
Solution 2. Putting x = 0 = y in the given equation f(x) + f(y) = f(x + y), we get
f(0) + f(0) = f(0 + 0)
or, 2f(0) = f(0)
or, f(0) = 0 (1)
Next, putting x = −y in the given equation, we get
f(x) + f(−x) = f(x −x)
or, f(x) + f(−x) = f(0)
or, f(x) + f(−x) = 0 [by (1)]
Hence, f(−x) = −f(x), so that f is odd.