#2**+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.

Bertie Apr 23, 2014

#1**+5 **

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

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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.......

CPhill Apr 23, 2014

#2**+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