Thank you for reporting, we will resolve it shortly
Q.
The co-efficient of $x^4$ in the expansion of $\left(1-x+2 x^2\right)^{12}$ is :
Binomial Theorem
Solution:
$\left(1- x +2 x ^2\right)^{12}=\left((1- x )+2 x ^2\right)^{12} $
$=(1- x )^{12}+{ }^{12} C _1(1- x )^{11} 2 x ^2+{ }^{12} C _2(1- x )^{10} 4 x ^4+\ldots \ldots $
$\text { required co-efficient of } x ^4= $
$\text { co-eff. of } x ^4 \text { in }(1- x )^{12}+24 \text { co-eff. of } x ^2 \text { in }(1- x )^{11}+4 \cdot{ }^{12} C _2 $
$={ }^{12} C _4+24 \cdot{ }^{11} C _2+4 \cdot{ }^{12} C _2={ }^{12} C _4+\frac{2 \cdot 12 \cdot 11 \cdot 10 \cdot 3}{1 \cdot 2 \cdot 3}+4 \cdot{ }^{12} C _2$
$\left.={ }^{12} C _4+6 \cdot{ }^{12} C _3+4 \cdot{ }^{12} C _2={ }^{12} C _4+2 \cdot{ }^{12} C _3+4 \cdot{ }^{12} C _3+{ }^{12} C _2\right)$
$={ }^{12} C _4+2 \cdot{ }^{12} C _3+4 \cdot{ }^{13} C _3={ }^{12} C _4+{ }^{12} C _3+{ }^{12} C _3+4 \cdot{ }^{13} C _3$
$={ }^{13} C _4+{ }^{13} C _3+{ }^{12} C _3+3 \cdot{ }^{13} C _3={ }^{14} C _4+3 \cdot{ }^{13} C _3+{ }^{12} C _3 \Rightarrow D$