Question

Arrange the expansion of ${(x^{1/2}+\frac{1}{2x^{1/4}})}^n$ in decreasing powers of x. Suppose the coefficients of the first three terms form an arithmetic progression. Then the number of terms in the expansion having integer powers of $x$ is:

• A. $1$
• B. $2$
• C. $3$
• D. more than 3

Answer 1

The correct option is C $3$

$S={(x^{1/2}+\frac{1}{2x^{1/4}})}^n$

$=\sum {{}^{n}C}_r.{(x}^{1/2}{)}^{n-r}.{\left(\frac{1}{2x^{1/4}}\right)}^r$

$=\sum {{}^{n}C}_r.{\left(\frac{1}{2}\right)}^r.x^{(n-r)/2}.x^{-r/4}$

$\Rightarrow S={\sum }_{r=0}^n{{}^{n}C}_r{\left(\frac{1}{2}\right)}^r{x}^{(2n-3r)/4}$

We have to expand in decreasing powers of $x.$

$\therefore S={{}^{n}C}_0{\left(\frac{1}{2}\right)}^0{x}^{n/2}+{{}^{n}C}_1{\left(\frac{1}{2}\right)}^1{x}^{(2n-3)/4}+{{}^{n}C}_2{\left(\frac{1}{2}\right)}^2{x}^{(2n-6)/4}+......$

First three terms form an $A.P.$

$={{}^{n}C}_0,{{}^{n}C}_1\left(\frac{1}{2}\right)\&{{}^{n}C}_2{\left(\frac{1}{2}\right)}^2$ from an $A.P.$

$\Rightarrow {{}^{n}C}_0+{{}^{n}C}_2{\left(\frac{1}{2}\right)}^2=2\left({{}^{n}C}_1\right).\frac{1}{2}$

$\Rightarrow 1+\frac{n(n-1)}{2}\times \frac{1}{4}=n$

$\Rightarrow (n-1)(n-8)=0$

$n\ne 1$ ($\because$ there are at least $3$ term)

$\Rightarrow n=8$

$\therefore S={\sum }_{r=0}^8{{}^{8}C}_r{\left(\frac{1}{2}\right)}^2{x}^{4-3r/4}$

$\Rightarrow 4-\frac{3r}{4}=K;$ Where $K\in I$

Now of such values of $r$ ranging from $0$ to $8$ are $:\to 0,4$ and $8$

$\therefore$ There will be three such terms.

Hence, the answer is $3.$