Question

The sum of the binomial coefficients of ${[2x+\frac{1}{x}]}^n$ is equal to $256$. The constant term in the expansion is-

• A. $1120$
• B. $2110$
• C. $1210$
• D. None of the above

Answer 1

Answer: (A)

$1120$

We know that Sum of binomial coeficients is:

${\Rightarrow }^n{C}_0{+}^n{C}_1{+}^n{C}_2+...{.}^n{C}_n=2^n$

Given that

$\Rightarrow 2^n=256$

$\Rightarrow n=8$

$\Rightarrow T_r={}^n{C}_r\left(2x{)}^{n-r}\right(\frac{1}{x}{)}^r$

${\Rightarrow }^8{C}_r(2x{)}^{8-r}{x}^{-r}$

${\Rightarrow }^8{C}_r(2{)}^{8-r}{x}^{8-2r}$

For contant term , the power of $x$ is zero.

Hence, $\Rightarrow 8-2r=0$

$\Rightarrow r=4$

$\Rightarrow T_4={}^8{C}_4(2{)}^{8-4}$

$\Rightarrow T_4={}^8{C}_4(2{)}^4$

$\Rightarrow T_4=1120$