An algebra problem by Aswad Hariri Mangalaeng

Forget the 0 so from the number -1 to the number 2014 has 2014 numbers, each pair of successive numbers (e.g -1 + 2, -3 + 4, ..., -2013 + 2014) is always equal to 1. Since 2014 numbers has 1007 pairs of successive numbers (2014/2) that each pair of successive numbers is equal to 1, the sum of terms will be 1007 x 1 = 1007!!!!

The voices in my head told me it was 1007.

we can write each term as sum of 1 and previous number
like 2=1+1 and 2014=1+2013
forget the 0 and total number of terms are 2014 and two terms result as single 1
Thus there will be 2014/2 terms of 1 i.e (1+1+1.............+1 (total 1007) )
hence answer will be 1007x1=1007

for even/odd numbers,the summation is $=\frac{n}{2}$ $(2a+(n-1)d)$

for even,n=1007;a=2;d=2

for,odd,n=1007;a=1;d=2

and then, $\text{bigger value-smaller value}\ =1007$

here we can see that odd nos. are being subtracted from the even nos.

therefore sum of all even nos. - sum of all odd nos.

formula : for sum off n odd nos . = n^2

formula for sum of n even nos. = n(n+1)

= putting the values we get = 1007(1007+1) - 1007*1007

= 1015056 - 1014049

= 1007 :-)

0-1+2-3+4-5+6-7+.................-2013+2014 since 0-1+2-3+4-5 can be written as (0-1)+(2-3)+(4-5)=-1-1-1=-3 this is the same as -((5+1)/2) if we do the sum up to -7, we get the answer as -4, which is -((7+1)/2). so you can see a pattern. In that sense, we can conclude that if we stop at -2013, that is if we do the sum till -2013, ie 0-1+2-3+4-5+................+2012-2013, this adds up to -((2013+1)/2)=-1007. The required sum is 0-1+2-3+4-5+6-7+................+2012-2013+2014=-1007+2014=1007.