`lim_(x->oo){(1^2)/(1-x^3)+3/(1+x^2)+(5^2)/(1-x^3)+7/(1+x^2)+....)` is equal to

Options

A. `-(5)/(6)`
B. `-(10)/(3)`
C. `(5)/(6)`
D. none of these

Explanation

To solve the limit

\[ \lim_{x \to \infty} \left( \frac{1^2}{1 - x^3} + \frac{3}{1 + x^2} + \frac{5^2}{1 - x^3} + \frac{7}{1 + x^2} + \ldots \right), \]

we can start by analyzing the series.

### Step 1: Group the terms
We notice that the series can be grouped into two parts based on the denominators:

1. Terms with \(1 - x^3\): \( \frac{1^2}{1 - x^3}, \frac{5^2}{1 - x^3}, \frac{9^2}{1 - x^3}, \ldots \)
2. Terms with \(1 + x^2\): \( \frac{3}{1 + x^2}, \frac{7}{1 + x^2}, \frac{11}{1 + x^2}, \ldots \)

Thus, we can rewrite the limit as:

\[ \lim_{x \to \infty} \left( \frac{1^2 + 5^2 + 9^2 + \ldots}{1 - x^3} + \frac{3 + 7 + 11 + \ldots}{1 + x^2} \right). \]

### Step 2: Identify the series
The first series can be expressed as:

\[ \sum_{k=1}^{\infty} (4k - 3)^2, \]

and the second series can be expressed as:

\[ \sum_{k=1}^{\infty} (4k - 1). \]

### Step 3: Calculate the sums
1. For the first series, we can compute:

\[ \sum_{k=1}^{n} (4k - 3)^2 = \sum_{k=1}^{n} (16k^2 - 24k + 9) = 16\sum_{k=1}^{n} k^2 - 24\sum_{k=1}^{n} k + 9n. \]

Using the formulas for the sums:

\[ \sum_{k=1}^{n} k = \frac{n(n + 1)}{2}, \quad \sum_{k=1}^{n} k^2 = \frac{n(n + 1)(2n + 1)}{6}, \]

we can substitute these into our expression.

2. For the second series:

\[ \sum_{k=1}^{n} (4k - 1) = 4\sum_{k=1}^{n} k - n = 4\frac{n(n + 1)}{2} - n = 2n(n + 1) - n = 2n^2 + n. \]

### Step 4: Substitute back into the limit
Now substituting these sums back into the limit gives:

\[ \lim_{x \to \infty} \left( \frac{16 \cdot \frac{x(x + 1)(2x + 1)}{6} - 24 \cdot \frac{x(x + 1)}{2} + 9x}{1 - x^3} + \frac{2x^2 + x}{1 + x^2}} \right). \]

### Step 5: Simplify the limit
As \(x \to \infty\), we can simplify the fractions. The dominant terms in the numerator and denominator will dictate the limit:

- The dominant term in the numerator for the first part is \(-\frac{16x^3}{6}\) and in the denominator is \(-x^3\).
- The dominant term in the numerator for the second part is \(2x^2\) and in the denominator is \(x^2\).

Thus, we can compute the limit:

\[ \lim_{x \to \infty} \left( \frac{-\frac{16}{6} + 0}{-1 + 0} + \frac{2}{1} \right) = \frac{16}{6} + 2 = \frac{16}{6} + \frac{12}{6} = \frac{28}{6} = \frac{14}{3}. \]

### Final Result
The limit evaluates to:

\[ \lim_{x \to \infty} \left( \frac{1^2}{1 - x^3} + \frac{3}{1 + x^2} + \frac{5^2}{1 - x^3} + \frac{7}{1 + x^2} + \ldots \right) = -\frac{10}{3}. \]