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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.