The coefficient of $x^{50}\text{ in }(1+x{)}^{41}{\left(1-x+x^2\right)}^{40}$ is 

$\begin{array}{l}E=(1+x){\left[(1+x)\left(1-x+x^2\right)\right]}^{40}\\ =(1+x){\left(1+x^3\right)}^{40}\\ T_{r+1}=(1+x){}^{40}{C}_r{\left(x^3\right)}^r\end{array}$

The power of x will be $3r\text{ or }3r+1$.

If $3r=50\text{ or }3r+1=50$ then both do not give integral

value of r and hence no term will be of the type $\lambda x^{50}$

Hence coefficient of $x^{50}\text{ is }0$.