Evaluate the limit (1^2-2^2+3^2-4^2+...+(2n-1)^2-(2n)^2)/(4n^2-3). n-->infinite

We notice that the summation from numerator could be
written as:

Sum [(2k - 1)^2 - (2k)^2], where k takes values
from 1 to n.

We'll expand the binomial and we'll
compute:

(2k - 1)^2 - (2k)^2 = 4k^2 - 4k + 1 -
4k^2

We'll eliminate like terms and we'll
get:

(2k - 1)^2 - (2k)^2 = 1 -
4k

Sum [(2k - 1)^2 - (2k)^2] = Sum 1 - Sum
4k

Sum 1 = 1+1+1+...+1 = n*1 =
n

Sum 4k = 4*Sum k =
4*(1+2+3+....+n)

We recognize inside brackets the sum of n
natural terms:

1+2+3+....+n =
n*(n+1)/2

4*Sum k =
4*n*(n+1)/2

4*Sum k = 2*n*(n+1) = 2n^2 +
2n

Now, we'll evaluate the
limit:

lim (n - 2n^2 - 2n)/(4n^2 - 3) = lim (-2n^2 -
n)/(4n^2 - 3)

Since the order of the numerator is matching
to the order of denominator, the limit is the ratio of the leading
coefficients:

lim (-2n^2 - n)/(4n^2 - 3) =
-2/4

lim (-2n^2 - n)/(4n^2 - 3) =
-1/2

The requested limit, if n approaches to
infinite, is: lim (-2n^2 - n)/(4n^2 - 3) = -1/2.