Question

In the expansion of ${(3^{\frac{-x}{4}}+3^{\frac{5x}{4}})}^n$, the sum of the binomial coefficients is $64$ and the term with the greatest binomial coefficient exceeds the third by $(n-1)th$ then find the value of $x$

• A. $\frac{\mathrm{6.5.4}}{\mathrm{1.2.3}}{3}^{\frac{3x}{4}}.3^{\frac{15x}{4}}$
• B. $\frac{\mathrm{6.5.4}}{\mathrm{1.2.3}}{3}^{-\frac{3x}{2}}.3^{\frac{15x}{4}}$
• C. $\frac{\mathrm{6.5.4}}{\mathrm{1.2.3}}{3}^{\frac{4x}{3}}.3^{\frac{15x}{4}}$
• D. $\frac{\mathrm{6.5.4}}{\mathrm{1.2.3}}{3}^{-\frac{3x}{4}}.3^{\frac{15x}{4}}$

Answer 1

Answer: (D)

$\frac{\mathrm{6.5.4}}{\mathrm{1.2.3}}{3}^{-\frac{3x}{4}}.3^{\frac{15x}{4}}$

Given the sum of the binomial coefficients in the expansion of ${(3^{\frac{-x}{4}}+3^{\frac{5x}{4}})}^n=64$

Then, substitute $3^{\frac{-x}{4}}=3^{\frac{5x}{4}}=1\therefore {(1+1)}^n=64\Rightarrow 2^n=64\therefore n=6$

we now that middle term has greatest binomial coefficient,

Here $n=6\therefore$ middle term$=(\frac{6}{2}+1)thterm=4thterm=T_4$ and given $T_4=(n-1){+T}_3\therefore T_{3+1}=(n-1)+T_{2+1}{{}^{6}C}_3{\left(3^{\frac{-x}{4}}\right)}^3{\left(3^{\frac{5x}{4}}\right)}^3=(6-1){{}^{6}C}_2{\left(3^{\frac{-x}{4}}\right)}^4{\left(3^{\frac{5x}{4}}\right)}^2\Rightarrow \frac{\mathrm{6.5.4}}{\mathrm{1.2.3}}{3}^{-\frac{3x}{4}}.3^{\frac{15x}{4}}$