Question

If sum of all the coefficients in the expansion of $(x^{3/2}+x^{1/3}{)}^n$ is 128, then the coefficient of $x^5$ is

• A. 35
• B. 45
• C. 7
• D. none of these

Answer 1

The correct option is A 35

Substituting $x=1$, we get the sum of the coefficients as
$(2{)}^n=128$
$\therefore n=7$
Hence writing the general term, we get
$T_{r+1}={}^7{C}_r{x}^{\frac{63-11r}{6}}$
Hence for the coefficient of $x^5$
$63-11r=6\left(5\right)$
$63-11r=30$
$33-11r=0$
$\therefore r=3$
Hence coefficient is ${}^7{C}_3$ $=35$.