If we put 0 in the middle and then look for 10 consecutive integers with 5 being positive and then other 5 being the negatives of those, it will equal 0.
So (-5)+(-4)+(-3)+(-2)+(-1)+0+1+2+3+4+5.
It is very easy to see this works by just paring up each number with its additive inverse.
So - 5 and 5 cancel out, 4 and -4 add up to 0 etc. And of course, 0 it the additive identity so it does not change the sum.
The sum of the squares of the three consecutive integers 11, 13, 15 = 515
36
7 + 9 + 11 = 27
11, 13, 15
The integers are -11 and -10.
The sum of the squares of the three consecutive integers 11, 13, 15 = 515
36
The integers are 11, 13 and 15.
7 + 9 + 11 = 27
11
-11 and -10
The integers are 10, 11 and 12.
(-11) + (-12) + -(13) = -36
39
They are: 10+11+12 = 33
They are 9, 10 and 11.
11, 12, and 13