Compute the sum\[ (a +(2n+1)d)^2- (a + (2n)d)^2 +(a + (2n-1)d)^2 - (a+(2n-2)d)^2 + \cdots + (a+d)^2 - a^2.\]

Answer 1

n
∑ k^2 = 1/6 n(n+1)(2n+1)
k=1

well, this is just the sum(k=a .. a+2n+1)
which is just sum(k=1 .. a+2n+1) - sum(k=1 .. a-1)