If the sum of the binomial coefficients of the expansion

Solution:

(a) 1120

Suppose $(r+1)$ th tem in the given expansion is independent of $x$.

Then, we have

$T_{r+1}={ }^{n} C_{r}(2 x)^{n-r}\left(\frac{1}{x}\right)^{r}$

$={ }^{n} C_{r} 2^{n-r} x^{n-2 r}$

For this term to be independent of $x$, we must have

$n-2 r=0$

$\Rightarrow r=n / 2$

$\therefore$ Required term $={ }^{n} C_{n / 2} 2^{n-n / 2}=\frac{n !}{[(n / 2) !]^{2}} 2^{n / 2}$

We know :

Sum of the given expansion $=256$ Thus,

we have

$2^{n} \cdot 1^{n}=256$

$\Rightarrow n=8$

$\therefore$ Required term $=\frac{8 !}{(4) !(4) !} 2^{4}=1120$