Question

Which term in the expansion of ${\left( {1 + x} \right)^p}{\left( {1 + \dfrac{1}{x}} \right)^q}$ is independent of x, where p, q are positive integers? What is the value of that term?

Answer 1

Hint: In this particular type of question use the concept that according to Binomial expansion, the expansion of, ${\left( {b + a} \right)^n} = \sum\limits_{r = 0}^n {{}^n{C_r}{b^{n - r}}{{\left( a \right)}^r}}$and later on use the concept that the independent term of x in the Binomial expansion of the given equation is constant term in the expansion of${\left( {1 + x} \right)^p}{\left( {1 + \dfrac{1}{x}} \right)^q}$ which we find out by equating the power of x to zero, so use these concepts to reach the solution of the question.

Complete step-by-step answer:
Given equation:
${\left( {1 + x} \right)^p}{\left( {1 + \dfrac{1}{x}} \right)^q}$
So first simplify this equation we have,
$\Rightarrow {\left( {1 + x} \right)^p}{\left( {1 + \dfrac{1}{x}} \right)^q} = {\left( {1 + x} \right)^p}{\left( {\dfrac{{x + 1}}{x}} \right)^q}$
$\Rightarrow {\left( {1 + x} \right)^p}{\left( {1 + \dfrac{1}{x}} \right)^q} = \dfrac{1}{{{x^q}}}{\left( {1 + x} \right)^{p + q}}$..................... (1)
So, first out constant term in the expansion of${\left( {1 + x} \right)^{p + q}}$.
As we know according to Binomial expansion, the expansion of
${\left( {b + a} \right)^n} = \sum\limits_{r = 0}^n {{}^n{C_r}{b^{n - r}}{{\left( a \right)}^r}}$
So, on comparing $b = 1,{\text{ }}a = x,{\text{ }}n = p + q$
$\Rightarrow {\left( {1 + x} \right)^{p + q}} = \sum\limits_{r = 0}^{p + q} {{}^{p + q}{C_r}{{\left( 1 \right)}^{p + q - r}}{{\left( x \right)}^r}} = \sum\limits_{r = 0}^{p + q} {{}^{p + q}{C_r}{{\left( x \right)}^r}}$, [as any power of 1 is always 1]
Now substitute this value in equation (1) we have,
$\Rightarrow {\left( {1 + x} \right)^p}{\left( {1 + \dfrac{1}{x}} \right)^q} = \dfrac{1}{{{x^q}}}\sum\limits_{r = 0}^{p + q} {{}^{p + q}{C_r}{{\left( x \right)}^r}}$
$\Rightarrow {\left( {1 + x} \right)^p}{\left( {1 + \dfrac{1}{x}} \right)^q} = \sum\limits_{r = 0}^{p + q} {{}^{p + q}{C_r}{{\left( x \right)}^{r - q}}}$
Now find out a constant term.
So, put the power of $x$in the expansion of ${\left( {1 + x} \right)^p}{\left( {1 + \dfrac{1}{x}} \right)^q}$ equal to zero.
$\Rightarrow r - q = 0$
$\Rightarrow r = q$
So the term independent of x in the expansion of ${\left( {1 + x} \right)^p}{\left( {1 + \dfrac{1}{x}} \right)^q}$ is ${}^{p + q}{C_r} = {}^{p + q}{C_q}$
Now as we know that ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ so we have,
$\Rightarrow {}^{p + q}{C_q} = \dfrac{{\left( {p + q} \right)!}}{{q!\left( {p + q - q} \right)!}} = \dfrac{{\left( {p + q} \right)!}}{{q!\left( p \right)!}}$
So this is the required answer.

Note – Whenever we face such types of questions the key concept we have to remember is that always recall the formula of the combination (i.e.${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$), then first simplify the given equation as above then expand the term using Binomial expansion as above then equating the power of x to zero as above, then simplify we will get the required answer.