Question

If the coefficient of first three terms in the expansion $\left(\sqrt{ x }+\frac{1}{2 \cdot \sqrt[4]{ x }}\right)^{ n }$ where $n \in N$ form an arithmetic progression, then find the number of terms in the expansion having integral powers of $x$.

Answer 1

$ T _{ r +1}={ }^{ n } C _{ r }(\sqrt{ x })^{ n - r } \cdot\left(\frac{1}{2 \cdot \sqrt[4]{ x }}\right)^{ r }$
Also, ${ }^{ n } C _0, \frac{{ }^{ n } C _1}{2}, \frac{{ }^{ n } C _2}{4}$ (in that order) are in A.P.
$\Rightarrow 2\left(\frac{{ }^{ n } C _1}{2}\right)={ }^{ n } C _0+\frac{{ }^{ n } C _2}{4} \Rightarrow n =8$
$\therefore T _{ r +1}=\frac{{ }^8 C _{ r }}{2^{ r }} \cdot( x )^{\frac{16-3 r }{4}} ; r =0,1,2,3,4,5,6,7,8$
But $\left(\frac{16-3 r}{4}\right)$ is an integer for $r=0,4,8$.