Find the sum of the series
1^2 – 2^2 + 3^2 – 4^2 + … - (2n)^2
How can I solve this?
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.
∑(1-4n) = -n(2n + 1)
And this is the correct answer.