Question

The sum of the binomial coefficients in the expansion of ${(x^{-3/4}+a{x}^{5/4})}^n$ lies between $200$ and $400$ and the term independent of $x$ equals $448$. The value of $a$ is

• A. $1$
• B. $2$
• C. $1/2$
• D. for no value of $a$

Answer 1

Answer: (B)

$2$

${(x^{-\frac{3}{4}}+a{x}^{5/4})}^n$ $={}^n{C}_0{\left(x^{-3/4}\right)}^n+{}^n{C}_1\left(x^{-3/4}{)}^{n-1}\right(a{x}^{5/4}{)}^1.......+{}^n{C}_n(a{x}^{5/4}{)}^n$

Put $x=1$

$(1+a{)}^n={}^n{C}_0+{}^n{C}_{1}a+{}^n{C}_2{a}^2+......+{}^n{C}_n{a}^n$

Put $a=1$

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

$200<2^n<400$

$[\because 2^8=256]$

$n=8$

$(x^{-3/4}+a{x}^{5/4}{)}^8$

$T_{r+1}={}^8{C}_r\left(x^{-3/4}{)}^{8-r}\right(a{x}^{5/4}{)}^r$

$\Rightarrow T_{3+1}={}^8{C}_3{a}^3=448$

$T_4{=}^8{C}_3{a}^3=448$

$\Rightarrow \frac{8\times 7\times 6}{3\times 2}=a^3=448\Rightarrow a^3=8$

$a=2$

$\left[\begin{array}{c}\because \frac{-3}{4}(8-r)+\frac{5r}{4}=0\\ -6+\frac{3r}{4}+\frac{5r}{4}=0\\ -6+2r=0\\ r=3\end{array}\right]$.