The general term is given by
Tr + 1 = 1024Cr \((5^{\frac{1}{2}})^{1024-r} (7^{\frac{1}{8}})^{r}\)
= 1024Cr 5 512 Cr5 512 −\(\frac{r}{2}-7^\frac{r}{8}\)
={ 1024Cr 5 512-r} x \(5^\frac{r}{2}\times7^\frac{r}{8}\)
= { 1024Cr 5 512-r} x \((5^4\times7)^\frac{r}{8}\)
Clearly, Tr + 1 will be an integer if \(\frac{r}{8}\)is an integer such that 0 ≤ r ≤ 1024 R is multiple of 8 satisfying 0 ≤ r ≤ 1024 r = 0, 8, 16, 24,…, 1024 r can assumes 129 values.
Hence, there are 129 integral terms in the expansion of \((5^{\frac{1}{2}} + 7^{\frac{1}{4}})^{1024}\)