In the expansion of ${\left(5^{\frac{{\displaystyle 1}}{{\displaystyle 2}}}+7^{\frac{{\displaystyle 1}}{{\displaystyle 8}}}\right)}^{1024}$, the number of integral terms is 

To find number of integral terms in the expansion of  ${\left(5^{\frac{{\displaystyle 1}}{{\displaystyle 2}}}+7^{\frac{{\displaystyle 1}}{{\displaystyle 8}}}\right)}^{1024}$

The general term $T_{r+1}$ in the expansion of  ${\left(5^{\frac{{\displaystyle 1}}{{\displaystyle 2}}}+7^{\frac{{\displaystyle 1}}{{\displaystyle 8}}}\right)}^{1024}$ is

$T_{r+1}{=}^{1024}{C}_r{\left(5^{\frac{{\displaystyle 1}}{{\displaystyle 2}}}\right)}^{1024-r}{\left(7^{\frac{{\displaystyle 1}}{{\displaystyle 8}}}\right)}^r$

$\begin{array}{l}{T}_{r+1}{=}^{1024}{C}_r{.5}^{\frac{1024-r}{2}{.7}^{\frac{r}{8}}}\\ {=}^{1024}{C}_r{5}^{512-r+\frac{r}{2}}{7}^{\frac{r}{8}}\\ {=}^{1024}{C}_r{\left(5\right)}^{512-r}.\left(5^{\frac{r}{2}}{.7}^{\frac{r}{8}}\right)\end{array}$

${=}^{1024}{C}_r{\left(5\right)}^{512-r}.{\left(5^4\times 7\right)}^{\frac{r}{8}}$

Clearly $T_{r+1}$ will be an integer, iff  $\frac{r}{8}$ is an integer such that

$0\le r\le 1024$

$\Rightarrow r$  0,8,16,24,…,1024

Take 8,16,24,…,1024

$0\le r\le 1024$

$\Rightarrow r=0,8,16,24,32,\cdots ,1024$

 Formula: $T_n=a+(n-1)d$

$\Rightarrow 1024=8+(n-1)8\Rightarrow 1024=8n\Rightarrow n=128$

Including 0, r can assume 129 values.