Question

In the expansion of $(3^{-x/4}+3^{5x/4}{)}^n$, the sum of binomial coefficient is 64 and term with the greatest bionomial coefficient term exceeds the third term by $(n-1)$ the value of x must be

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

Answer: (A)

$0$

${(3^{-x/4}+3^{5x/4})}^n={\sum }_{r=0}^n{}^n{C}_r{\left(3^{-x/4}\right)}^{n-r}.{\left(3^{5x/4}\right)}^{n-r}$

We know that sum of binomial coefficient $=2^n$

$\Rightarrow 2^n=6^4$

$\Rightarrow n=6$

Third term, $T_3={}^6{C}_2{\left(3^{-x/4}\right)}^4.{\left(3^{5x/4}\right)}^2$

$\Rightarrow 15\times 3^{(-x+5x/2)}=15.\left(3^{3x/2}\right)$

Term with greatest binomial coefficient will be the middle term ; i.e. $T_4$

$\Rightarrow T_4={}^6{C}_3{\left(3^{-x/4}\right)}^3.{\left(3^{5x/4}\right)}^3$

$={20.3}^{3x}$

Given that, ${20.3}^{3x}-{15.3}^{3x/2}=6-1=5.$

Solving the above equation by taking $\left(3^{3x/2}\right)$ as $y.$

We get $x=0.$

Hence, the answer is $0.$