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Skip to 0 minutes and 10 secondsIn 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 secondsAnd 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 secondsNow, 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 secondsLet 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 secondsNow, 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 secondsAgain, 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 secondAnd 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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Precalculus: the Mathematics of Numbers, Functions and Equations

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