1. ## Problem in series

Find the sum of the series

1^2 – 2^2 + 3^2 – 4^2 + … - (2n)^2

How can I solve this?

2. You can find out the solution here.
http://www.mathhelpforum.com/math-he...10-series.html

3. As we know,
nth term = a + (n - 1)d

so: –(4n-1) = 1– 4n

There I’ve confusion.

If LHS=RHS, what’s they mean?

We need to do summation of (1-4n) only? Or nth term is (1-4n)?

But in the arithmetic series we know summation of the series is Sn = n/2 [2a + (n - 1)d]

By following arithmetic formulae I didn’t get the correct answer.

But If
n
(1-4n) = -n(2n + 1)
r=1

And this is the correct answer.