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

# $\underset{x\to \mathrm{\infty }}{lim} \left\{\frac{{1}^{2}}{1-{x}^{3}}+\frac{3}{1+{x}^{2}}+\frac{{5}^{2}}{1-{x}^{3}}+\frac{7}{1+{x}^{2}}+\dots \right\}$ equals

1. A

$-\frac{5}{6}$

2. B

$-\frac{10}{3}$

3. C

$\frac{5}{6}$

4. D

none of these

Register to Get Free Mock Test and Study Material

+91

Verify OTP Code (required)

### Solution:

We have,

$\begin{array}{l}\underset{x\to \mathrm{\infty }}{lim} \left\{\frac{{1}^{2}}{1-{x}^{3}}+\frac{3}{1+{x}^{2}}+\frac{{5}^{2}}{1-{x}^{3}}+\frac{7}{1+{x}^{2}}+\dots \right\}\\ =\underset{x\to \mathrm{\infty }}{lim} \left\{\frac{{1}^{2}+{5}^{2}+{9}^{2}+\dots }{1-{x}^{3}}+\frac{3+7+11+\dots }{1+{x}^{2}}\right\}\\ =\underset{x\to \mathrm{\infty }}{lim} \left\{\frac{\sum _{k=1}^{x} \left(4k-3{\right)}^{2}}{1-{x}^{3}}+ \frac{\sum _{k=1}^{x}\left(4k-1\right)}{1+{x}^{2}}\right\}\\ \begin{array}{l}=\underset{x\to \mathrm{\infty }}{lim} \left\{\frac{16\sum _{k=1}^{x} {k}^{2}-24\sum _{k=1}^{x} k+\sum _{k=1}^{x} 9}{1-{x}^{3}}+\frac{2x\left(x+1\right)-x}{1+{x}^{2}}\right\}\\ \underset{x\to \mathrm{\infty }}{lim} \left\{\frac{16x\left(x+1\right)\left(2x+1\right)-12x\left(x+1\right)+9x}{1-{x}^{3}}\right\}\\ =\frac{-32}{6}+2=\frac{-10}{3}\end{array}\end{array}$

Register to Get Free Mock Test and Study Material

+91

Verify OTP Code (required)