(b) Here , a power of $2$, where as the power of $ 7$ is $\frac{1}{8} = {2^{ – 3}}$

Now first term $^{1024}{C_0}{\left( {{5^{1/2}}} \right)^{1024}} = {5^{512}}$ (integer)

And after $8$ terms, the $9^{th}$ term ${ = ^{\,\,\,1024}}{C_8}{({5^{1/2}})^{1016}}{({7^{1/8}})^8}$ = an integer

Again, $17^{th}$ term =$^{1024}{C_{16}}{({5^{1/2}})^{1008}}{({7^{1/8}})^{16}}$
= An integer.

Continuing like this, we get an $A.P.$ $1, 9, 17, …., 1025,$

because $1025^{th}$ term = the last term in the expansion

$ = {\,^{1024}}{C_{1024}}{\left( {{7^{1/8}}} \right)^{1024}} = {7^{128}}$(an integer)

If $n$ is the number of terms of above $A.P$. we have

$1025 = {T_n} = 1 + (n – 1)8\,\,\, \Rightarrow n = 129$.