MCQ
The coefficient of $x^{2012}$ in $\frac{1+x}{\left(1+x^2\right)(1-x)}$ is
- ✓$1$
- B$2011$
- C$2012$
- D$2013$
We have, $\frac{1+x}{\left(1+x^2\right)} \frac{(1-x)}{}$
$\frac{1+x}{\left(1+x^2\right)(1-x)}=\frac{x}{1+x^2}+\frac{1}{2\left(1+x^2\right)}$
$+\frac{1}{2(1-x)}$
$=x\left(1+x^2\right)^{-1}+\frac{1}{2}\left(1+x^2\right)^{-1}+\frac{1}{2}(1-x)^{-1}$
Coefficient of $x^{2012}$ in
$\left[x\left(1+x^2\right)^{-1}+\frac{1}{2}\left(1+x^2\right)^{-1}+\frac{1}{2}(1-x)^{-1}\right]$ is
$=0+\frac{1}{2}+\frac{1}{2}=1$
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$1.$ The probability that $x_1+x_2+x_3$ is odd, is $x _1+ x _2+ x _3$
$(A)$ $\frac{29}{105}$ $(B)$ $\frac{53}{105}$ $(C)$ $\frac{57}{105}$ $(D)$ $\frac{1}{2}$
$2.$ The probability that $x_1, x_2, x_3$ are in an arithmetic progression, is
$(A)$ $\frac{9}{105}$ $(B)$ $\frac{10}{105}$ $(C)$ $\frac{11}{105}$ $(D)$ $\frac{7}{105}$
Give the answer question $1$ and $2.$