If

$(20)^{19}+2(21)(20)^{18}+3(21)^{2}(20)^{17}+\ldots+20(21)^{19}=k(20)^{19}$,

then $k$ is equal to ___________.

$\begin{aligned} &(20)^{19}+2(21)(20)^{18}+3(21)^2(20)^{17} \\ & \quad+\ldots \ldots+20(21)^{19}=k(20)^{19} \\\\ & \Rightarrow(20)^{19}\left[1+2\left(\frac{21}{20}\right)+3\left(\frac{21}{20}\right)^2+\ldots+20\left(\frac{21}{20}\right)^{19}\right]=k(20)^{19} \\\\ & \Rightarrow k=1+2\left(\frac{21}{20}\right)+3\left(\frac{21}{20}\right)^2+\ldots+20\left(\frac{21}{20}\right)^{19} ..........(i)\end{aligned}$ <br/><br/>Now, <br/><br/>$\begin{aligned} k\left(\frac{21}{20}\right)=\left(\frac{21}{20}\right) & +2\left(\frac{21}{20}\right)^2+3\left(\frac{21}{20}\right)^3 +\ldots+20\left(\frac{21}{20}\right)^{20}\end{aligned}$ ........(ii) <br/><br/>On subtracting Equation (ii) from Equation (i), we get <br/><br/>$$ \begin{aligned} & k\left(\frac{-1}{20}\right)=1+\frac{21}{20}+\left(\frac{21}{20}\right)^2+\ldots \ldots+\left(\frac{21}{20}\right)^{19}-20\left(\frac{21}{20}\right)^{20} \\\\ & \Rightarrow k\left(\frac{-1}{20}\right)=\frac{1\left(\left(\frac{21}{20}\right)^{20}-1\right)}{\frac{21}{20}-1}-20\left(\frac{21}{20}\right)^{20} \\\\ & \Rightarrow k\left(\frac{-1}{20}\right)=20\left(\left(\frac{21}{20}\right)^{20}-1\right)-20\left(\frac{21}{20}\right)^{20} \\\\ & =20\left(\frac{21}{20}\right)^{20}-20-20\left(\frac{21}{20}\right)^{20} \\\\ & \Rightarrow k\left(\frac{-1}{20}\right)=-20 \\\\ & \Rightarrow k=400 \end{aligned} $$