Plus Advent Calendar Door #2: Shake to solve

Sometimes real progress in maths comes when you find a way of looking at a problem in two different ways. Here is a great example of this.

handshake

Suppose you have $n+1$ people in a room and each person shakes hands with each other person once. How many handshakes do you get in total? The first person shakes hands with $n$ other people, the second shakes hands with the $n-1$ remaining people, the third shakes hands with $n-2$ remaining people, etc, giving a total of

$n+(n-1)+(n-2)+...+2+1$ handshakes.

But we can also look at this in another way: each person shakes hands with $n$ others and there are $n+1$ people, giving $n \times (n+1)$ handshakes. But this counts every handshake twice, so we need to divide by 2, giving a total of

  \[ \frac{n \times (n+1)}{2} \]    

handshakes.

Putting these two arguments together, we have just come up with the formula for summing the first $n$ integers and we’ve proved that it is correct:

  \[ n+(n-1)+(n-2)+...+2+1 = \frac{n \times (n+1)}{2}. \]    

This puzzle is inspired by content on our sister site Wild Maths, which encourages students to explore maths beyond the classroom and is designed to nurture mathematical creativity. The site is aimed at 7 to 16 year-olds, but open to all. It provides games, investigations, stories and spaces to explore, where discoveries are to be made. Some have starting points, some a big question and others offer you a free space to investigate.

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Comments

I think this is how Gauss managed to sum the integers 1 to 100 in record time when he was a schoolboy as 1+2+....+100 is the same as 50*101.