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# Arrange the expansion of ${\left( {{x^{\dfrac{1}{2}}} + \dfrac{1}{{2{x^{\dfrac{1}{2}}}}}} \right)^n}$ in the decreasing powers of $x$. Suppose the coefficients of the first three terms are in 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 $3$

Last updated date: 19th Jun 2024
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Hint: Here we can write the general binomial expansion of ${\left( {{x^{\dfrac{1}{2}}} + \dfrac{1}{{2{x^{\dfrac{1}{2}}}}}} \right)^n}$ which we can write by:
${\left( {{x^{\dfrac{1}{2}}} + \dfrac{1}{{2{x^{\dfrac{1}{2}}}}}} \right)^n}$$= \sum\limits_{r = 0}^n {{}^n{C_r}} {({x^{\dfrac{1}{2}}})^{n - r}}{\left( {\dfrac{1}{{{2^{\dfrac{1}{4}}}}}} \right)^r}$
And then we can write the coefficients of the first three terms as $2b = a + c$ if the first three terms are $a,b,c$ and then we can find the number of terms easily with integral powers of $x$.

Complete step by step solution:
Here we are given to write the expansion of ${\left( {{x^{\dfrac{1}{2}}} + \dfrac{1}{{2{x^{\dfrac{1}{2}}}}}} \right)^n}$
So we must know that the general binomial expansion is given by:
${\left( {a + b} \right)^n}$$= \sum\limits_{r = 0}^n {{}^n{C_r}} {(a)^{n - r}}{\left( b \right)^r}$
So similarly we can write the expansion of the given binomial expansion and write it as:
${\left( {{x^{\dfrac{1}{2}}} + \dfrac{1}{{2{x^{\dfrac{1}{2}}}}}} \right)^n}$$= \sum\limits_{r = 0}^n {{}^n{C_r}} {({x^{\dfrac{1}{2}}})^{n - r}}{\left( {\dfrac{1}{{2{x^{\dfrac{1}{4}}}}}} \right)^r}$
Now we can simplify and write it as:
${\left( {{x^{\dfrac{1}{2}}} + \dfrac{1}{{2{x^{\dfrac{1}{2}}}}}} \right)^n}$$= \sum\limits_{r = 0}^n {{}^n{C_r}} {x^{\dfrac{{n - r}}{2}}}{\left( {\dfrac{1}{2}} \right)^r}{x^{\dfrac{{ - r}}{4}}}$$= \sum\limits_{r = 0}^n {{}^n{C_r}} {\left( {\dfrac{1}{2}} \right)^r}{(x)^{\dfrac{{n - r}}{2} - \dfrac{r}{4}}}$
So we get the simplified form as:
${\left( {{x^{\dfrac{1}{2}}} + \dfrac{1}{{2{x^{\dfrac{1}{2}}}}}} \right)^n}$$= \sum\limits_{r = 0}^n {{}^n{C_r}} {\left( {\dfrac{1}{2}} \right)^r}{(x)^{\dfrac{{2n - 3r}}{4}}}$$ - - - - - (1)$
Now we have to expand the decreasing power of $x$ so we can substitute the values of $r = 0,1,2,....$ and so on.
So we get:
${\left( {{x^{\dfrac{1}{2}}} + \dfrac{1}{{2{x^{\dfrac{1}{2}}}}}} \right)^n}$$= {}^n{C_0}{\left( {\dfrac{1}{2}} \right)^0}{(x)^{\dfrac{{2n - 3\left( 0 \right)}}{4}}} + {}^n{C_1}{\left( {\dfrac{1}{2}} \right)^1}{(x)^{\dfrac{{2n - 3\left( 1 \right)}}{4}}} + {}^n{C_2}{\left( {\dfrac{1}{2}} \right)^2}{(x)^{\dfrac{{2n - 3\left( 2 \right)}}{4}}} + ......... {\left( {{x^{\dfrac{1}{2}}} + \dfrac{1}{{2{x^{\dfrac{1}{2}}}}}} \right)^n}$$ = {}^n{C_0}{(x)^{\dfrac{n}{2}}} + {}^n{C_1}\left( {\dfrac{1}{2}} \right){(x)^{\dfrac{{2n - 3}}{4}}} + {}^n{C_2}{\left( {\dfrac{1}{2}} \right)^2}{(x)^{\dfrac{{2n - 6}}{4}}} + .........$$- - - - - \left( 2 \right) So we get this as the expansion of the term in the decreasing powers of the variable x Now we are also given that the first three coefficients are in arithmetic progression which means these three terms have the common difference between them. We know that if a,b,c are in AP then 2b = a + c Hence we can write from the expansion of equation \left( 2 \right) that: ${}^n{C_0},{}^n{C_1}\left( {\dfrac{1}{2}} \right),{}^n{C_2}{\left( {\dfrac{1}{2}} \right)^2}$ are in AP, hence we can write that: 2{}^n{C_1}\left( {\dfrac{1}{2}} \right) = {}^n{C_0} + {}^n{C_2}{\left( {\dfrac{1}{2}} \right)^2} \\ {}^n{C_1} = 1 + \dfrac{{n!}}{{2!\left( {n - 2} \right)!}}.\dfrac{1}{4} \\ {}^n{C_1} = 1 + \dfrac{{n\left( {n - 1} \right)}}{8} Hence we can write that: n = 1 + \dfrac{{n\left( {n - 1} \right)}}{8} \\ \left( {n - 1} \right) - \dfrac{{n\left( {n - 1} \right)}}{8} = 0 \\ \left( {n - 1} \right)\left( {1 - \dfrac{n}{8}} \right) = 0 \\ We can further write it as: - \dfrac{1}{8}\left( {n - 1} \right)\left( {n - 8} \right) = 0 \\ \left( {n - 1} \right)\left( {n - 8} \right) = 0 \\ n = 1{\text{ or 8}} \\ But we know that n cannot be 1 because we are told that there are at least 3 terms. Hence we can say that n = 8 Substitute n = 8 in equation (1) we get: {\left( {{x^{\dfrac{1}{2}}} + \dfrac{1}{{2{x^{\dfrac{1}{2}}}}}} \right)^8}$$ = \sum\limits_{r = 0}^8 {{}^8{C_r}} {\left( {\dfrac{1}{2}} \right)^r}{(x)^{4 - \dfrac{{3r}}{4}}}$
Hence know that power of the $x$ has to be integer which means that $4 - \dfrac{{3r}}{4} \in Z$
Therefore $\dfrac{{3r}}{4} \in Z$
So $r$ has to be multiple of $4$ so that denominator and numerator cancel each other and we get the integer.
So as $r \in \left[ {0,8} \right]$
Hence we can say that $r = 0,4,8$

Hence we can say that there are $3$ values of $r$
Option C) is the correct answer.

Note:
Whenever the student is given to expand the terms of the bracket of the form ${\left( {a + b} \right)^n}$ then we must know that its expansion can be written in the form of ${\left( {a + b} \right)^n}$$= \sum\limits_{r = 0}^n {{}^n{C_r}} {(a)^{n - r}}{\left( b \right)^r}$
Therefore according to this expansion we can solve for what we are required to find.