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

### 2 Answers | Add Yours

The expression [(1^2 - 2^2 + 3^2 - 4^2 +...+ (2n-1)^2 - (2n)^2)] /(4n^2-3) can be rewritten as

[(1^2 - 2^2) + (3^2 - 4^2) +...+ ((2n-1)^2 - (2n)^2))] /(4n^2-3)

=> [(1 -2)(1 + 2) + (3-4)(3 + 4)+...+ - (4n)] /(4n^2-3)

=> [-3 - 7 - 11 -... - (4n)] /(4n^2-3)

The numerator is an AP, with the first term -3 and the common difference -4. -4n = -3 + (t - 1)*(-4)

=> t = 1 + (4n - 3)/4 = (4n + 1)/4

The sum of t terms is

(t/2)[2a + (t-1)*d]

=> ((4n + 1)/8)*[-6 + (4n - 3)*-1]

=> ((4n + 1)/8)*[-6 - 4n + 3)]

=> (4n + 1)(-4n - 3)/8

=> (-16n^2 - 16n - 3)/8

=> -2n^2 - 2n - 3/8

lim n--> inf.[(-2n^2 - 2n - 3/8)/(4n^2 - 3)]

substituting n = inf., gives the indeterminate form inf./inf.

So use l'Hopital's rule and substitute the numerator and denominator with their derivatives.

=> lim n--> inf.[(-4n - 2)/(8n)]

Again use l'Hopital's rule

=> lim n--> inf.[(-4/8)]

**The value of the required limit is -1/2**

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.**