Question

If $\sum _{k=1}^{19}\frac{(-2{)}^k}{k!(19-k)!}=\frac{-\lambda }{19!}$ then find $\lambda$.

Answer 1

$\begin{array}{c}{\sum }_{k=1}^{19}\frac{{\left(-2\right)}^k}{k!\left(19-k\right)!}=\frac{-\lambda }{19!}\\ \Rightarrow {\sum }_{k=1}^{19}\frac{19!{\left(-2\right)}^{{}^k}}{k!\left(19-k\right)!}19!=\frac{-\lambda }{19!}\\ \Rightarrow {\sum }_{k=1}^{19}\frac{{}^{19}{C}_k{\left(-2\right)}^k{\left(1\right)}^{19-k}}{19!}=\frac{-\lambda }{19!}\\ \Rightarrow \frac{{\left(-2H\right)}^{19}{-}^{19}{C}_0}{19!}=-\frac{\lambda }{19!}\\ \Rightarrow -1-1=-\lambda \\ \Rightarrow \lambda =2\end{array}$