limx→∞ 121−x3+31+x2+521−x3+71+x2+… equals

A
$- \frac{5}{6}$
B
$- \frac{10}{3}$
C
$\frac{5}{6}$
D
none of these

Solution:

We have,

$\begin{array}{l} lim_{x \to \infty} \{\frac{1^{2}}{1 - x^{3}} + \frac{3}{1 + x^{2}} + \frac{5^{2}}{1 - x^{3}} + \frac{7}{1 + x^{2}} + \dots\} \\ = lim_{x \to \infty} \{\frac{1^{2} + 5^{2} + 9^{2} + \dots}{1 - x^{3}} + \frac{3 + 7 + 11 + \dots}{1 + x^{2}}\} \\ = lim_{x \to \infty} \{\frac{\sum_{k = 1}^{x} (4 k - 3)^{2}}{1 - x^{3}} + \frac{\sum_{k = 1}^{x} (4 k - 1)}{1 + x^{2}}\} \\ \begin{array}{l} = lim_{x \to \infty} \{\frac{16 \sum_{k = 1}^{x} k^{2} - 24 \sum_{k = 1}^{x} k + \sum_{k = 1}^{x} 9}{1 - x^{3}} + \frac{2 x (x + 1) - x}{1 + x^{2}}\} \\ lim_{x \to \infty} \{\frac{16 x (x + 1) (2 x + 1) - 12 x (x + 1) + 9 x}{1 - x^{3}}\} \\ = \frac{- 32}{6} + 2 = \frac{- 10}{3} \end{array} \end{array}$

$lim_{x \to \infty} \{\frac{1^{2}}{1 - x^{3}} + \frac{3}{1 + x^{2}} + \frac{5^{2}}{1 - x^{3}} + \frac{7}{1 + x^{2}} + \dots\}$ equals

Solution:

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