If the sum of the coefficients of all the positive powers of $\mathrm{x}$, in the binomial expansion of ${\left(x^n+\frac{2}{x^5}\right)}^7$ is 939 , then the sum of all the possible integral values of $\mathrm{n}$ is :

$$\begin{array}{rl} & {(x^n+\frac{2}{x^5})}^7=\sum _{r=0}^7{ }^7{C}_r\left(x^n\right)\cdot {\left(\frac{2}{x^5}\right)}^r\\ & =\sum _{r=0}^7{ }^7{C}_r\cdot 2^r\cdot x^{7n-nr-5r}\\ & \therefore { }^7{C}_0\cdot 2^0+{ }^7{C}_1\cdot 2^1+{ }^7{C}_2\cdot 2^2+{ }^7{C}_3\cdot 2^3+{ }^7{C}_4\cdot 2^4\\ & =939\\ & \therefore r=4\\ & \because 7n-nr-5r=0\\ & \text{ and }r=4\text{ then }n>\frac{20}{3}\\ & \text{ and }r\text{ should not be }5\\ & \therefore n<\frac{25}{2}\\ & \therefore \text{ Possible values of }n\text{ are }7,8,9,10,11,12\\ & \therefore \text{ Sum of integral value of }n=57\end{array}$$