Skip to 0 minutes and 10 seconds In exercise 3, we are dealing with the famous formula stating that the sum of the integers from 1 to n equals n times n plus 1 over 2. Here, n is an integer, and is greater than 1. So let us call P of n this proposition. And we check the validity of P of n for n equal to 1, 2, and 3. Now, for n equal 1, P of 1 states that 1 equals 1 times 2 over 2. Yes, it’s true. For n equal to 2, P of 2 states that 1 plus 2, which is 3, equals 2 times 3 over 2. Well, it’s true.

Skip to 1 minute and 19 seconds And for n equal to 3, P of 3 states that sum– 1 plus 2 plus 3, which is actually equal to 6, equals 3 times 4 over 2, which is also 6. So it is true.

Skip to 1 minute and 40 seconds Now, there is a story behind this formula. It seems that the teacher wanted to punish Gauss when he was at school and asked him to count the integers from 1 to 100. He wanted to give him a difficult and boring task. However, Gauss immediately answered, after a few seconds, answered that the sum of the integers from 1 to 100 equals 50 times 101. Well, how did he get this? Well, simply, he proved this formula and he applied it to n equal to 100. The application to n equal to 100 gives 100 times 101 divided by 2, which is exactly 50 times 101. So he probably proved this formula by what is called Gauss method.

Skip to 2 minutes and 38 seconds Let us write– let us call S of n the sum of the integers from 1 to n. So 1 plus 2 plus plus plus plus n minus 1, plus n.

Skip to 2 minutes and 58 seconds Now, let us write again this sum by reversing the order of the terms. So we begin with n, and then we’ve got n minus 1, plus plus, plus 2 plus 1. And the reason is that if we sum the terms on each column, we get n plus 1. So here we get n plus 1. Here we get n plus 1.

Skip to 3 minutes and 34 seconds Again, here we get n plus 1. And here we get n plus 1. How many times does n plus 1 up here? Well, exactly n times. So there are n terms in this sum. So the sum of these terms, which is twice the sum of that we are considering, is equal to n times n plus 1.

Skip to 4 minutes and 1 second And this is also 2 times the sum that you are considering. The sum of the terms from 1 to n. So we get that 2 times S of n equals n times n plus 1. So if we divide by 2, we immediately obtain that the sum of the integers from 1 to n equals n times n plus 1 over 2. This ends exercise 3 and also this step in practice. See you in the next step in practice.

# Discussion in practice: Gauss sum

When Gauss, who later became a famous mathematician, was still in school, his teacher asked him to sum the integers from 1 to 100. The teacher probably wanted to punish him with a boring task. However, after a few seconds he answered: 5050.

In the next exercise we see how Gauss got this result. Can you think of any alternative ways to reach the conclusion?

### Exercise 3.

For \(n\in\mathbb N,\, n\ge 1,\) let \(P(n)\) be the proposition \[P(n):\quad 1+2+\cdots + n=\dfrac{n(n+1)}2.\]

1) Verify that \(P(1), P(2)\) and \(P(3)\) are true.

2) Prove the result by mean of Gauss method as follows:

i) Call \(S(n)=1+2+\cdots +n\). Write in two subsequent rows the sum \(S(n)\) by reversing the order of its terms: \[(a)\quad S(n)=1+2+\cdots+n\] \[(b)\quad S(n)=n+(n-1)+\cdots +1\] Notice that \(1+n=2+(n-1)=3+(n-2)=…\).

ii) Sum term by term on the columns in (a) and (b): you get \(2S(n)=\cdots\)

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