Question

Arrange the expansion of ${\left( {{x^{\dfrac{1}{2}}} + \dfrac{1}{{2{x^{\dfrac{1}{4}}}}}} \right)^n}$ in decreasing power 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

First, we have to arrange the expansion in decreasing power of x by using the binomial theorem formula.
Binomial theorem: It is the method of expanding an expression which has been raised to any finite power.
${\left( {x + y} \right)^n} = {x^n} + \left( {{}^n{C_1}} \right){x^{n - 1}}y + \left( {{}^n{C_2}} \right){x^{n - 2}}{y^2} + ..... + \left( {{}^n{C_{n - 1}}} \right)x{y^{n - 1}} + y$
$\Rightarrow {\left( {x + y} \right)^n} = \sum\limits_{r = 0}^n {\left( {{}^n{C_r}} \right){x^{n - r}}{y^r}}$
A combination is an arrangement of objects, without repetition, and order not being important.
${}^n{C_r} = c\left( {n,r} \right) = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
A sequence is called an arithmetic progression if and only if the difference of any term from its preceding term is constant. Property of an A.P (arithmetic progression): If a, b, and c are in A. P., then$2b = a + c$.

Complete step by step answer:
We have,
$S = {\left( {{x^{\dfrac{1}{2}}} + \dfrac{1}{{2{x^{\dfrac{1}{4}}}}}} \right)^n}$
By binomial theorem,
$\Rightarrow S = \sum\limits_{r = 0}^n {\left( {{}^n{C_r}} \right){{\left( {{x^{\dfrac{1}{2}}}} \right)}^{n - r}}{{\left( {\dfrac{1}{{2{x^{\dfrac{1}{4}}}}}} \right)}^r}}$
Simplify it.
$\Rightarrow S = \sum\limits_{r = 0}^n {\left( {{}^n{C_r}} \right){{\left( {\dfrac{1}{2}} \right)}^r}{{\left( x \right)}^{\dfrac{{\left( {n - r} \right)}}{2}}}{{\left( x \right)}^{\dfrac{{ - r}}{4}}}}$
Let us simplify x terms.
$\Rightarrow S = \sum\limits_{r = 0}^n {\left( {{}^n{C_r}} \right){{\left( {\dfrac{1}{2}} \right)}^r}{{\left( x \right)}^{\dfrac{{\left( {2n - 3r} \right)}}{4}}}}$………1)
Now, we have to expand in decreasing powers of x by putting values of r.
$\therefore S = {}^n{C_0}{\left( {\dfrac{1}{2}} \right)^0}{\left( x \right)^{\dfrac{n}{2}}} + {}^n{C_1}{\left( {\dfrac{1}{2}} \right)^1}{\left( x \right)^{\dfrac{{\left( {2n - 3} \right)}}{4}}} + {}^n{C_2}{\left( {\dfrac{1}{2}} \right)^2}{\left( x \right)^{\dfrac{{\left( {2n - 6} \right)}}{4}}} + ....$
In this above expansion first three terms form an arithmetic progression.
$\Rightarrow {}^n{C_0},{}^n{C_1}\left( {\dfrac{1}{2}} \right),$ ${}^n{C_2}{\left( {\dfrac{1}{2}} \right)^2}$ form an A.P.
By the property of arithmetic progression which is explained in the hint.
$\Rightarrow {}^n{C_0} + {}^n{C_2}{\left( {\dfrac{1}{2}} \right)^2} = 2\left( {{}^n{C_1}} \right)\left( {\dfrac{1}{2}} \right)$
Use the combination formula.
$\Rightarrow 1 + \dfrac{{n\left( {n - 1} \right)}}{2} \times \dfrac{1}{4} = n$
Simplify it.

$\Rightarrow 1 + \dfrac{{n\left( {n - 1} \right)}}{8} = n$

To simplify it again, we have to do cross multiplication.
$\Rightarrow 1 + {n^2} - n = 8n$
$\Rightarrow 1 + {n^2} - n - 8n = 0$
$\Rightarrow {n^2} - 9n + 1 = 0$
To find the values of n, let us find the factors.
We get,
$\Rightarrow \left( {n - 1} \right)\left( {n - 8} \right) = 0$
But $n \ne 1$($\because$ there are at least three terms).
So, n=8.
We will put this value in (1).
$\Rightarrow S = \sum\limits_{r = 0}^8 {\left( {{}^8{C_r}} \right){{\left( {\dfrac{1}{2}} \right)}^r}{{\left( x \right)}^{\dfrac{{\left( {2(8) - 3r} \right)}}{4}}}}$
Simplify it.
$\Rightarrow S = \sum\limits_{r = 0}^8 {\left( {{}^8{C_r}} \right){{\left( {\dfrac{1}{2}} \right)}^r}{{\left( x \right)}^{\dfrac{{\left( {16 - 3r} \right)}}{4}}}}$
$\Rightarrow S = \sum\limits_{r = 0}^8 {\left( {{}^8{C_r}} \right){{\left( {\dfrac{1}{2}} \right)}^r}{{\left( x \right)}^{4 - \dfrac{{3r}}{4}}}}$
Let us find the number of terms in the expansion having integer powers of x.
$\Rightarrow 4 - \dfrac{{3r}}{4} = k$ ; $k \in {\rm I}$
Substitute the values of r from 0 to 8.
Then we get integer values of k at $r = 0,4,8$
Hence, there will be three such terms.
So, the answer is 3.

So, the correct answer is “Option C”.

Note: Applications of the binomial theorem.
A Binomial theorem is a powerful tool of expansion which has application in algebra & probability.
The binomial theorem is used in forecast services. The disaster forecast also depends upon the use of binomial theorems.