Arrange the expansion of left(x^{1 / 2}+frac{ | vedclass

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Question

(c)

We have, $\left(x^{1 / 2}+\frac{1}{2 x^{1 / 4}}\right)^n$

$T_{r+1}={ }^n C_r\left(x^{1 / 2}\right)^{n-r} \frac{1}{\left(2 x^{\frac{1}{4}}\ri

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(c)

We have, $\left(x^{1 / 2}+\frac{1}{2 x^{1 / 4}}ight)^n$

$T_{r+1}={ }^n C_r\left(x^{1 / 2}ight)^{n-r} \frac{1}{\left(2 x^{\frac{1}{4}}ight)^r}$

$T_{r+1}={ }^n C_r x{ }^{n-r}-{ }^r 4 \cdot 2^{-r}$

$T_{r+1}={ }^n C_r \frac{2 n-3 r}{4} \cdot 2^{-r}=\frac{{ }^n C_r}{2^r} x^{\frac{2 n-3 r}{4}}$

Given $T_1, T_2, T_3$ are in AP.

$	herefore 2 T_2=T_1+T_3$

$\frac{2^n C_1}{2}={ }^n C_0+\frac{{ }^n C_2}{2^2}$

$\Rightarrow { }^n C_1={ }^n C_0+\frac{n(n-1)}{2 \cdot 2^2}$

$\Rightarrow \quad n=1+\frac{n(n-1)}{8}$

$\Rightarrow \quad n-1=\frac{n(n-1)}{8}$

$\Rightarrow n-1=0 	ext { or } n=8$

When $n=8$,

$\frac{2 n-3 r}{4}=\frac{16-3 r}{4}$ is integer

If $\quad r=0,4,8$

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

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