The value of the sum ${ }^{50}{C}_1-(1+\frac{1}{2}){}^{50}{C}_2+(1+\frac{1}{2}+\frac{1}{3}){}^{50}{C}_3-\mathrm{.......}-(1+\frac{1}{2}+\frac{1}{3}+\mathrm{......}+\frac{1}{50}){}^{50}{C}_{50}=\frac{1}{k}$ then $\frac{k}{10}$ must be ….

$$\begin{gathered} { }^{50}{C}_1-(1+\frac{1}{2}){}^{50}{C}_2+(1+\frac{1}{2}+\frac{1}{3}){}^{50}{C}_3\mathrm{......}(1+\frac{1}{2}+\frac{1}{3}+\mathrm{.....}\frac{1}{50}){}^{50}{C}_{50}{ \\ }^{50}{C}_1\underset{0}{\overset{1}{\int }}1dx{}^{50}{C}_2\underset{0}{\overset{1}{\int }}(1+x)dx{+}^{50}{C}_3\underset{0}{\overset{1}{\int }}(1+x+x^2)dx{\mathrm{.....}}^{50}{C}_{50}\underset{0}{\overset{1}{\int }}(1+x+x^2+\mathrm{.....}+x^{49})dx \\ {=}^{50}{C}_1\underset{0}{\overset{1}{\int }}\frac{1-x}{1-x}dx-{}^{50}{C}_2\underset{0}{\overset{1}{\int }}\frac{1-x^2}{1-x}dx+{}^{50}{C}_3\underset{0}{\overset{1}{\int }}\frac{1-x^3}{1-x}dx\mathrm{.........}{}^{50}{C}_{50}\underset{0}{\overset{1}{\int }}\frac{1-x^{50}}{1-x}dx \\ =\underset{0}{\overset{1}{\int }}\frac{{ }^{50}{C}_1{-}^{50}{C}_2+{\mathrm{......}}^{50}{C}_{50}}{1-x}dx-\underset{0}{\overset{1}{\int }}\frac{{(}^{50}{C}_1{-}^{50}{C}_2{x}^2{\mathrm{.......}}^{50}{C}_{50})dx}{1-x} \\ {(1-x)}^{50}={}^{50}{C}_0-{}^{50}{C}_{1}x++{}^{50}{C}_2{x}^2-{}^{50}{C}_3{x}^3\mathrm{......}+{}^{50}{C}_{50}{x}^{50} \\ {}^{50}{C}_1-{}^{50}{C}_2+{}^{50}{C}_3\mathrm{.......}-{}^{50}{C}_{50}=1 \\ {}^{50}{C}_{1}x-{}^{50}{C}_2{x}^2+{}^{50}{C}_3{x}^3\mathrm{..........}{}^{50}{C}_{50}{x}^{50}=1-{(1-x)}^{50} \\ =\underset{0}{\overset{1}{\int }}\frac{1}{1-x}dx-\underset{0}{\overset{1}{\int }}\frac{1-{(1-x)}^{50}}{(1-x)}dx=\underset{0}{\overset{1}{\int }}{(1-x)}^{49}dx=\frac{1}{50}{/}_{\raisebox{1ex}{$K$}\!\left/ \!\raisebox{-1ex}{$10$}\right.=5}^{K=50} \end{gathered}$$