Question

Prove that there is no term involving ${{x}^{2}}$ in the expression of ${{\left( 2{{x}^{2}}-\dfrac{3}{x} \right)}^{11}}$, where $x\ne 0$.

Answer 1

Hint: In this question, we will use binomial theorem to expand the given expression and then check powers of $x$ in all the terms involved in expansion and check if ${{x}^{2}}$ is present or not.

Complete step-by-step answer:
Firstly, we will use binomial theorem to write expanded form of ${{\left( 2{{x}^{2}}-\dfrac{3}{x} \right)}^{11}}$.
Binomial theorem states that,
For any positive integer $n$, the ${{n}^{th}}$ power of the sum of two real numbers $a$ and $b$ may be expressed as the sum of $n+1$ terms as given below:
${{\left( a+b \right)}^{n}}=\sum\limits_{r=0}^{n}{{}^{n}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}}$
$\Rightarrow {{\left( a+b \right)}^{n}}={}^{n}{{C}_{0}}{{a}^{n}}+{}^{n}{{C}_{1}}{{a}^{n-1}}{{b}^{1}}+{}^{n}{{C}_{2}}{{a}^{n-2}}{{b}^{2}}+\cdots +{}^{n}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}+\cdots +{}^{n}{{C}_{n}}{{b}^{n}}$.
Where, ${}^{n}{{C}_{r}}=\dfrac{\left| \\!{\underline {\, n \,}} \right. }{\left| \\!{\underline {\, r \,}} \right. \times \left| \\!{\underline {\, n-r \,}} \right. }$.
Here, using binomial theorem in ${{\left( 2{{x}^{2}}-\dfrac{3}{x} \right)}^{11}}$, we have,
$a=2{{x}^{2}},\,b=-\dfrac{3}{x}\,\text{and}\,n=11$
Therefore,
${{\left( 2{{x}^{2}}-\dfrac{3}{x} \right)}^{11}}=\sum\limits_{r=0}^{11}{{}^{11}{{C}_{r}}{{\left( 2{{x}^{2}} \right)}^{11-r}}{{\left( -\dfrac{3}{x} \right)}^{r}}}$
Distributing power in numerator and denominator of $\dfrac{-3}{x}$ and multiplying powers of $2{{x}^{2}}$, we get,
${{\left( 2{{x}^{2}}-\dfrac{3}{x} \right)}^{11}}=\sum\limits_{r=0}^{11}{{}^{11}{{C}_{r}}2{{x}^{2\left( 11-r \right)}}\left( -\dfrac{{{3}^{r}}}{{{x}^{r}}} \right)}$
Applying distributive law in power of $x$, we get,
${{\left( 2{{x}^{2}}-\dfrac{3}{x} \right)}^{11}}=\sum\limits_{r=0}^{11}{{}^{11}{{C}_{r}}2{{x}^{22-2r}}\left( -\dfrac{{{3}^{r}}}{{{x}^{r}}} \right)}$
Separating $x$, we get,
${{\left( 2{{x}^{2}}-\dfrac{3}{x} \right)}^{11}}=\sum\limits_{r=0}^{11}{{}^{11}{{C}_{r}}2\left( -{{3}^{r}} \right)\dfrac{{{x}^{22-2r}}}{{{x}^{r}}}}$
Writing all exponential powers of $x$ in numerator, we get,
$\begin{aligned} & {{\left( 2{{x}^{2}}-\dfrac{3}{x} \right)}^{11}}=\sum\limits_{r=0}^{11}{{}^{11}{{C}_{r}}2\left( -{{3}^{r}} \right){{x}^{22-2r-r}}} \\\ & =\sum\limits_{r=0}^{11}{{}^{11}{{C}_{r}}2\left( -{{3}^{r}} \right){{x}^{22-3r}}} \\\ \end{aligned}$
Here, the value of $r$ is an integer, which lies from 0 to 11.
Therefore, from above equation, we can say that, terms of $x$ in the expansion of ${{\left( 2{{x}^{2}}-\dfrac{3}{x} \right)}^{11}}$ are:
${{x}^{22-3\times 0}},\,{{x}^{22-3\times 1}},\,{{x}^{22-3\times 2}},\,{{x}^{22-3\times 3}},\,{{x}^{22-3\times 4}},\,{{x}^{22-3\times 5}},\,{{x}^{22-3\times 6}},\,{{x}^{22-3\times 7}},\,{{x}^{22-3\times 8}},{{x}^{22-3\times 9}},\,{{x}^{22-3\times 10}}$ and ${{x}^{22-3\times 11}}$
$={{x}^{22-0}},\,{{x}^{22-3}},\,{{x}^{22-6}},\,{{x}^{22-9}},\,{{x}^{22-12}},\,{{x}^{22-15}},\,{{x}^{22-18}},\,{{x}^{22-21}},\,{{x}^{22-24}},{{x}^{22-27}},\,{{x}^{22-30}}$ and ${{x}^{22-33}}$
$={{x}^{22}},\,{{x}^{19}},\,{{x}^{16}},\,{{x}^{13}},\,{{x}^{10}},\,{{x}^{7}},\,{{x}^{4}},\,{{x}^{1}},\,{{x}^{-2}},{{x}^{-5}},\,{{x}^{-8}}$and ${{x}^{-11}}$
Hence, we can see that, there is no term involving ${{x}^{2}}$ in expansion of expression ${{\left( 2{{x}^{2}}-\dfrac{3}{x} \right)}^{11}}$.
Hence proved.

Note: This question can also be done by actually expanding the whole expression with all its coefficients. But it must be noted that we only need to check if ${{x}^{2}}$ is present or not and don’t need its coefficient. So, full expansion is not needed.