
Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion 
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March 22nd, 2019, 03:26 PM  #1 
Newbie Joined: Mar 2019 From: Granada, Spain Posts: 2 Thanks: 0  1 + 2 + ... + n
Hello When reading "A Mathematical Bridge" from Stephen Hewson I came upon the following problem: https://ibb.co/pr4rDVB The book shows you different ways of solving it and I'm ok with all of them but this one: https://ibb.co/0Z9PkRw How is the lefthand side exactly rearranged? Why does it matter if n is even or odd? Basically, what is the followed process? Thank you very much for reading. I hope you can solve my question. P.S.: Please excuse me for my possible spelling/grammar mistakes. 
March 22nd, 2019, 04:00 PM  #2  
Senior Member Joined: Aug 2012 Posts: 2,414 Thanks: 755  Quote:
1 + 2 + 3 + 4 + 5 + 6 + 7 We add the smallest and largest number" 1 + 7 = 8. Now we add the next smallest and next largest: 2 + 6 = 8 Then 3 + 5 = 8. But now because 7 is odd, we are left with a single number in the center, which only gets counted once. So the total is 8, times the number of times 8 shows up, plus 4. How many times does 8 show up? n/2  1, it looks to me. You have to think about this a little. And why 8? It's 1 more than n, right? Again you need to stop and think at each step to see what's going on. So we have (n/2  1) x (n+1) + 4. For n = 7 we have 3 x 8 + 4 = 28. Which is 7x8/2, so that checks. But what if n is even? Say we have n = 8. So we have 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8. Now the pairs are 1 + 8 = 9, 2 + 7 = 9, 3 + 6 = 9, and 4 + 5 = 9. That's 4x9 = 36, and again that matches 8x9/2. But note that with an even number of terms, the last step is a sum of a pair; but with an odd number of terms, the last step is a solitary single number in the middle. It's like pairing off partners at a dance. If there's an odd number of people, there must be someone left out. But if there's an even number, eveyone gets a partner. Last edited by Maschke; March 22nd, 2019 at 04:05 PM.  
March 22nd, 2019, 04:01 PM  #3 
Global Moderator Joined: May 2007 Posts: 6,835 Thanks: 733 
It doesn't matter. The simplest proof (observed by 10 year old Gauss). 1+ 2+ 3+......+n n+ (n1)+(n2)+....+1 __________________________ (n+1)+(n+1)+(n+1).+(n+1)=n(n+1). Divide by 2 because you have 2 copies. 
March 22nd, 2019, 04:17 PM  #4 
Senior Member Joined: Oct 2015 From: Greece Posts: 138 Thanks: 8 
Gauss thought of that when he was quite young. He noticed that if you pair up these numbers like this (1, 100), (2, 99), (3 , 98) and so on you always get a sum of 101, 50 times. So the answer should be 50 times 101. Notice that his task was to find the sum of the first 100 numbers (his teacher was lazy and gave this assignment to all his students so time would pass keeping the children busy). Since you know some algebra, the best way to make you understand this is to show you using algebra, what Gauss was actually thinking. Explanation: Let's describe the sum x of the first 100 numbers using algebra. $\displaystyle 1+2+...+99+100 = x $ Now if I write this backwards it would also be the same sum, right? $\displaystyle 100+99+...+2+1 = x $ Now let's add these equations together: $\displaystyle 1 + 2+...+99+100 = x \\ 100+99+...+2+1 = x \\ (+) \\ 101+101+...+101+101 = 2x $ We have 101 100 times so: $\displaystyle 100 \cdot 101 = 2x \Leftrightarrow x = \frac{100 \cdot 101}{2} = 50 \cdot 101 $ Now for exercise try doing the same thing in general using the sum of the first n numbers and try to prove why $\displaystyle 1+2+3+...+n = \frac{n(n+1)}{2} $ Last edited by skipjack; March 24th, 2019 at 12:05 AM. 
March 23rd, 2019, 12:47 AM  #5 
Newbie Joined: Mar 2019 From: Granada, Spain Posts: 2 Thanks: 0 
Thank you very much everybody. My doubts have been resolved. Best wishes, 

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