Question 456820
Did you try to do this your self? Just do the math:

If a = first integer, then a+1, and a+2 are the other consecutive integers. The average of the squares is:

{{{(a^2 + (a+1)^2 + (a+2)^2)/3}}} = {{{a^2 + 2a + 5/3}}}

The square of the average is:

{{{((a+a+1+a+2)/3)^2}}} = {{{a^2+2a+1}}}

Hence, the difference is 5/3 - 1 = 2/3.

If the integers didn't increase by 1 and instead increased by b, so like instead of 3,4,5, it was 3, 3+b, 3+2b, for any b (if b = 2, then 3, 5, 7), then the difference is {{{2b^2/3}}}.

if b = 1, then this reduces to 2/3, which we got earlier.