Question

If there is term ${{x}^{2r}}$ in ${{\left( x+\dfrac{1}{{{x}^{2}}} \right)}^{n-3}}$ , then
(a) n – 2r is a positive integral multiple of 3.
(b) n – 2r is even
(c) n – 2r is odd
(d) None of these

Answer 1

First of all, we will understand what is the binomial expansion and define the general form of the binomial expansion. Then we will apply binomial expansion on the given binomial ${{\left( x+\dfrac{1}{{{x}^{2}}} \right)}^{n-3}}$ and try to find the general term of the binomial. We will assume that the variable is one such term is ${{x}^{2r}}$. Thus, we can equate the power of ${{x}^{2r}}$, i.e. 2r and the power of the variable in the general term and get a relation. Thus, based on the relation, we can choose one of the options.

Complete step-by-step solution:
A binomial, as the name suggests, is a polynomial with two terms. The binomial theorem is used to get the expression when a binomial is multiplied with itself n number of times.
Let (a + b) be a binomial. Suppose we multiply this binomial with itself n number of times. The resultant expression will be as follows:
$\Rightarrow$ (a + b)(a + b)(a + b)…. n times.
$\Rightarrow {{\left( a+b \right)}^{n}}$
Thus, according to the binomial expansion theorem, the expression ${{\left( a+b \right)}^{n}}$ can be expanded as follows:
$\Rightarrow {{\left( a+b \right)}^{n}}=\left( \begin{matrix} n \\\ 0 \\\ \end{matrix} \right){{a}^{n}}{{b}^{0}}+\left( \begin{matrix} n \\\ 1 \\\ \end{matrix} \right){{a}^{n-1}}{{b}^{1}}+\left( \begin{matrix} n \\\ 2 \\\ \end{matrix} \right){{a}^{n-2}}{{b}^{2}}+...+\left( \begin{matrix} n \\\ n \\\ \end{matrix} \right){{a}^{0}}{{b}^{n}}$
Where $\left( \begin{matrix} n \\\ r \\\ \end{matrix} \right){{=}^{n}}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$
Thus, the general term (r + 1)th of the expression will be ${{T}_{r+1}}=\left( \begin{matrix} n \\\ r \\\ \end{matrix} \right){{a}^{n-r}}{{b}^{r}}$ where r is a positive integer such that $0\le r\le n$.
The expression given to us is ${{\left( x+\dfrac{1}{{{x}^{2}}} \right)}^{n-3}}$. This is a binomial with a = x and b = $\dfrac{1}{{{x}^{2}}}$ and n = n – 3.
Thus, we will use binomial expansion theorem to expand ${{\left( x+\dfrac{1}{{{x}^{2}}} \right)}^{n-3}}$.

n-3 \\\ 0 \\\ \end{matrix} \right){{\left( x \right)}^{n-3}}{{\left( \dfrac{1}{{{x}^{2}}} \right)}^{0}}+\left( \begin{matrix} n-3 \\\ 1 \\\ \end{matrix} \right){{\left( x \right)}^{n-4}}{{\left( \dfrac{1}{{{x}^{2}}} \right)}^{1}}+\left( \begin{matrix} n-3 \\\ 2 \\\ \end{matrix} \right){{\left( x \right)}^{n-5}}{{\left( \dfrac{1}{{{x}^{2}}} \right)}^{2}}+...+\left( \begin{matrix} n-3 \\\ n-3 \\\ \end{matrix} \right){{\left( x \right)}^{0}}{{\left( \dfrac{1}{{{x}^{2}}} \right)}^{n-3}}$$ Where $\left( \begin{matrix} n-3 \\\ s \\\ \end{matrix} \right){{=}^{n}}{{C}_{s}}=\dfrac{\left( n-3 \right)!}{s!\left( n-s-3 \right)!}$ And the general (s + 1)th of the expression will be ${{T}_{s+1}}=\left( \begin{matrix} n-3 \\\ s \\\ \end{matrix} \right){{\left( x \right)}^{n-3-s}}{{\left( \dfrac{1}{{{x}^{2}}} \right)}^{s}}$, where s is a positive integer such that $0\le s\le n-3$. $\begin{aligned} & \Rightarrow {{T}_{s+1}}=\left( \begin{matrix} n-3 \\\ s \\\ \end{matrix} \right){{\left( x \right)}^{n-3-s}}{{\left( x \right)}^{-2s}} \\\ & \Rightarrow {{T}_{s+1}}=\left( \begin{matrix} n-3 \\\ s \\\ \end{matrix} \right){{\left( x \right)}^{n-3-3s}} \\\ \end{aligned}$ Let us assume that the (s + 1)th has the variable ${{x}^{2r}}$. $\Rightarrow c{{x}^{2r}}=\left( \begin{matrix} n-3 \\\ s \\\ \end{matrix} \right){{\left( x \right)}^{n-3-3s}}$ , where c is some constant. Thus, since the bases are same, we can equate the powers. $\Rightarrow $ 2r = n – 3 – 3s $\Rightarrow $ n – 2r = 3(s + 1) Thus, we can say that n – 2r is a positive integral multiple of 3. **Hence, option (a) is the correct option.** **Note:** Students are advised to be careful while using the binomial expansion theorem as it only applies when binomials are multiplied by itself. If there is a product of multiple binomials, it cannot be used.