Compute the sum \[ (a +(2n+1)d)^2- (a + (2n)d)^2 +(a + (2n-1)d)^2 - (a+(2n-2)d)^2 + \cdots + (a+d)^2 - a^2.\]
Using difference of squares, the sum works out to 2(a + (n + 1)d).
Can you please give some hints on how you got this answer? I get that you use difference of squares, but I couldn't find the last term. Thanks
