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1+2+3+4+5+6+7+8+9.........+48+49+50+49+48+47.....+5+4+3+2+1

 Apr 23, 2014

Best Answer 

 #2
avatar+890 
+8

Here's an alternative method.

Fold the 49+48+...+2+1 underneath the 1+2+...+48+49 and you have

$$\begin{array}{l}\quad 1 \;+ \;2+...+48+49+50\\+49+48+...+2 \;+1\end{array}$$

Now add 'columnwise' 1+49, 2+48, and so on, (the 50 on the top line is left by itself),  and you can see that you finish up with 50 50's giving a total of 2500.

 Apr 23, 2014
 #1
avatar+103049 
+5

Re: 1+2+3+4+5+6+7+8+9.........+48+49+50+49+48+47.....+5+4+3+2+1

----------------------------------------------------------------------------------------------------------------

This looks "tricky," but it's not....we're just adding the first 49 positive integers twice, and then adding 50 to that.

The sum of the first n positive integers is given by (n)(n+1)/2

So, twice that sum is given by 2*[(n)(n+1)/2] = (n)(n+1)

So, twice the sum of the first 49 integers is just (49)(50) and adding 50 to that, we have

(49)(50) + 50   =  50 (49 + 1) = 50(50) = 2500.......

 Apr 23, 2014
 #2
avatar+890 
+8
Best Answer

Here's an alternative method.

Fold the 49+48+...+2+1 underneath the 1+2+...+48+49 and you have

$$\begin{array}{l}\quad 1 \;+ \;2+...+48+49+50\\+49+48+...+2 \;+1\end{array}$$

Now add 'columnwise' 1+49, 2+48, and so on, (the 50 on the top line is left by itself),  and you can see that you finish up with 50 50's giving a total of 2500.

Bertie Apr 23, 2014
 #3
avatar+103049 
0

Thanks, Bertue.........you must have heard the old story about Gauss using EXACTLY that method to find the sum of the 1st 100 digits as an classroom assignment when he was 10 years old.

I much prefer your method to mine!!!

 Apr 23, 2014

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