For $\mathrm{x}=\frac{5}{7}$, if ${\mathrm{t}}_{\mathrm{k}}$ is the first negative term in the expansion of ${\left(1+\mathrm{x}\right)}^{\frac{7}{5}},$ then ${\mathrm{t}}_1+{\mathrm{t}}_2+\dots \dots +{\mathrm{t}}_{\mathrm{k}}=$

$\mathrm{x}=\frac{5}{7}, {(1+\mathrm{x})}^{7/5}$

First negative term $={\mathrm{T}}_{\left[\frac{\mathrm{p}}{\mathrm{q}}\right]+3}={\mathrm{T}}_{\left[\frac{7}{5}\right]+3}$

K=4                           $={\mathrm{T}}_4$

$$\begin{gathered} {(1+\mathrm{x})}^{7/5}=1+\frac{7}{5}\mathrm{x}+\frac{{\displaystyle \frac{7}{5}}\left({\displaystyle \frac{7}{5}}-1\right)}{2!}{\mathrm{x}}^2+\frac{{\displaystyle \frac{7}{5}}\left({\displaystyle \frac{7}{5}}-1\right) \left({\displaystyle \frac{7}{5}}-2\right)}{3!} {\mathrm{x}}^3+..... \\ {\mathrm{T}}_1+{\mathrm{T}}_2+{\mathrm{T}}_3+{\mathrm{T}}_4=1+\frac{7}{5}\mathrm{x}+\frac{\left({\displaystyle \frac{7}{5}}\right) \left({\displaystyle \frac{2}{5}}\right)}{2}{\mathrm{x}}^2+\frac{\left({\displaystyle \frac{7}{5}}\right) \left({\displaystyle \frac{2}{5}}\right) \left({\displaystyle \frac{-3}{5}}\right)}{6}{\mathrm{x}}^3 \\ =1+\frac{7}{5} \left(\frac{5}{7}\right)+\left(\frac{7}{5}\right) \left(\frac{1}{5}\right)\frac{25}{49}-\left(\frac{7}{5}\right) \left(\frac{1}{5}\right) \left(\frac{1}{5}\right) {\left(\frac{5}{7}\right)}^3 \\ =1+1+\frac{1}{7}-\frac{1}{49}=\frac{98+7-1}{49}=\frac{104}{49} \end{gathered}$$