Question

The term independent of x in the expansion of ${{\left( {{x}^{2}}-\dfrac{3\sqrt{3}}{{{x}^{3}}} \right)}^{10}}$ is?

Answer 1

Use the formula of expression of ${{\left( a+x \right)}^{n}}$ as ${{\left( a+x \right)}^{n}}{{=}^{n}}{{C}_{0}}{{a}^{0}}{{x}^{n}}{{+}^{n}}{{C}_{1}}{{a}^{1}}{{x}^{n-1}}{{+}^{n}}{{C}_{2}}{{a}^{2}}{{x}^{n-2}}+..........{{+}^{n}}{{C}_{n}}{{a}^{n}}{{x}^{0}}$
Where general term ${{(r+1)}^{th}}$ ${{T}_{r+1}}{{=}^{n}}{{C}_{r}}{{a}^{r}}{{x}^{n-r}}$
And the term ‘independent of x’ means which term will not contain i.e. power of ‘x’ in that term is 0. Use the above result and condition to solve the problem.

Complete step by step answer:
As we know the expansion of the ${{\left( a+x \right)}^{n}}$from the binomial theorem can be given as:
${{\left( a+x \right)}^{n}}{{=}^{n}}{{C}_{0}}{{a}^{0}}{{x}^{n}}{{+}^{n}}{{C}_{1}}{{a}^{1}}{{x}^{n-1}}{{+}^{n}}{{C}_{2}}{{a}^{2}}{{x}^{n-2}}+..........{{+}^{n}}{{C}_{n}}{{a}^{n}}{{x}^{0}}$ ………………………. (i)
And the general term of the expansion of the above series is given as
${{T}_{r+1}}{{=}^{n}}{{C}_{r}}{{a}^{r}}{{x}^{n-r}}$…………… (ii)
Where by putting r = 0, 1, 2, 3………….n, we get the terms of the expansion of ${{\left( a+x \right)}^{n}},{{T}_{r+1}}$ is the
${{(r+1)}^{th}}$ term of the expansion.
Now, we have the expansion ${{\left( {{x}^{2}}-\dfrac{3\sqrt{3}}{{{x}^{3}}} \right)}^{10}}$, and we need to determine the term independent of ‘x’. So, let ${{(r+1)}^{th}}$term of the expansion of the expression ${{\left( {{x}^{2}}+\left( \dfrac{-3\sqrt{3}}{{{x}^{3}}} \right) \right)}^{10}}$ will be independent of ‘x’.
So, ${{(r+1)}^{th}}$term can be calculated from the equation (ii) as
${{T}_{r+1}}{{=}^{n}}{{C}_{r}}{{a}^{r}}{{x}^{n-r}}$
Replace ‘a’ by $'{{x}^{2}}'$ and ‘x’ by
$\left( \dfrac{-3\sqrt{3}}{{{x}^{3}}} \right)$
In the above expression; and put n = 10 as well. So, we get
${{T}_{r+1}}{{=}^{10}}{{C}_{r}}{{\left( {{x}^{2}} \right)}^{r}}{{\left( \dfrac{-3\sqrt{3}}{{{x}^{3}}} \right)}^{10-r}}$ ………… (iii)
Now, we can use property of surds as
${{\left( {{m}^{n}} \right)}^{p}}={{m}^{np}}$ ……… (iii)
${{\left( \dfrac{m}{n} \right)}^{a}}=\dfrac{{{m}^{a}}}{{{n}^{a}}}$……………………. (iv)
$\dfrac{{{m}^{a}}}{{{m}^{b}}}={{m}^{a-b}}$………………. (v)
So, we can replace ${{\left( {{x}^{2}} \right)}^{r}}$ in the expression (iii) by ${{x}^{2r}}$ using the equation (iii) and replace
${{\left( \dfrac{-3\sqrt{3}}{{{x}^{3}}} \right)}^{10-r}}\Rightarrow {{\dfrac{\left( -3\sqrt{3} \right)}{{{\left( {{x}^{3}} \right)}^{10-r}}}}^{10-r}}$
Using the property (iv). Hence, we get equation (iii) as
${{T}_{r+1}}{{=}^{10}}{{C}_{r}}{{x}^{2r}}\dfrac{{{\left( -3\sqrt{3} \right)}^{10-r}}}{{{\left( {{x}^{3}} \right)}^{10-r}}}$
Replace ${{\left( {{x}^{3}} \right)}^{10-r}}\Rightarrow {{x}^{3\left( 10-r \right)}}$using the property (iii), so, we get
$\begin{aligned} & {{T}_{r+1}}{{=}^{10}}{{C}_{r}}{{x}^{2r}}\dfrac{{{\left( -3\sqrt{3} \right)}^{10-r}}}{{{x}^{3}}^{\left( ^{10-r} \right)}} \\\ & \Rightarrow {{T}_{r+1}}{{=}^{10}}{{C}_{r}}{{x}^{2r}}\dfrac{{{\left( -3\sqrt{3} \right)}^{10-r}}}{{{x}^{3}}^{0-3r}} \\\ & \Rightarrow {{T}_{r+1}}{{=}^{10}}{{C}_{r}}x\dfrac{{{x}^{2r}}}{{{x}^{30-3r}}}{{\left( -3\sqrt{3} \right)}^{10-r}} \\\ \end{aligned}$
Now, use the property (v) with the expression $\dfrac{{{x}^{2r}}}{{{x}^{30-3r}}}$ and so, we get

& {{T}_{r+1}}{{=}^{10}}{{C}_{r}}{{x}^{\left( 2r-\left( 30-3r \right) \right)}}{{\left( -3\sqrt{3} \right)}^{10-r}} \\\ & \Rightarrow {{T}_{r+1}}{{=}^{10}}{{C}_{r}}{{x}^{\left( 2r-30+3r \right)}}{{\left( -3\sqrt{3} \right)}^{10-r}} \\\ \end{aligned}$$ $\Rightarrow {{T}_{r+1}}{{=}^{10}}{{C}_{r}}{{x}^{\left( 5r-30 \right)}}{{\left( -3\sqrt{3} \right)}^{10-r}}$…………….. (vi) Now, we know that we have to determine the term which is independent of ‘x’ in the expansion of the given expression. So, in other words, we can say that the power of ‘x’ should be ‘0’ for getting a term independent of ‘x’. hence, the value of (5r – 30) i.e. power of ‘x’, should be ‘0’. Hence, on equating (5r – 30) to ‘0’, we get $\begin{aligned} & 5r-30=0 \\\ & \Rightarrow 5r=30 \\\ & \Rightarrow r=\dfrac{30}{5}=6 \\\ \end{aligned}$ Hence, put r = 6 in the expression (vi) to get the term independent of ‘x’. so, we get $$\begin{aligned} & {{T}_{6+1}}={{T}_{7}}{{=}^{10}}{{C}_{6}}{{x}^{0}}{{\left( -3\sqrt{3} \right)}^{10-6}}, \\\ &\Rightarrow {{T}_{7}}{{=}^{10}}{{C}_{6}}{{\left( -3\sqrt{3} \right)}^{4}} \\\ \end{aligned}$$ Now, use $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ to solve the expression $^{10}{{C}_{6}}$, where n! = 1.2.3.4…….n. so, we get $\begin{aligned} &\Rightarrow {{T}_{7}}=\dfrac{10!}{4!6!}\times {{\left( -3\sqrt{3} \right)}^{4}} \\\ &\Rightarrow {{T}_{7}}=\dfrac{10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}{4\times 3\times 2\times 1\times 6\times 5\times 4\times 3\times 2\times 1}\times {{\left( -3\sqrt{3} \right)}^{4}} \\\ &\Rightarrow {{T}_{7}}=10\times 3\times 7{{\left( -3\sqrt{3} \right)}^{4}} \\\ &\Rightarrow {{T}_{7}}=210\times {{\left( -3\sqrt{3} \right)}^{4}}=210\times {{\left( {{\left( -3\sqrt{3} \right)}^{2}} \right)}^{2}} \\\ &\Rightarrow {{T}_{7}}=210\times 729=153090 \\\ \end{aligned}$ Hence, ${{7}^{th}}$ term of the given expression will be independent of ‘x’ and can be given as 153090. **So, the correct answer is “153090”.** **Note:** One may expand ${{\left( a+n \right)}^{n}}$ as following: $${{\left( a+x \right)}^{n}}{{=}^{n}}{{C}_{0}}{{a}^{0}}{{x}^{n}}{{+}^{n}}{{C}_{1}}{{a}^{1}}{{x}^{n-1}}{{+}^{n}}{{C}_{2}}{{a}^{2}}{{x}^{n-2}}+..........{{+}^{n}}{{C}_{n}}{{a}^{n}}{{x}^{0}}$$ So, don’t confuse the expansion mentioned above and given in the solution, both are the same. One may calculate the whole expansion of ${{\left( {{x}^{2}}-\dfrac{3\sqrt{3}}{{{x}^{3}}} \right)}^{10}}$ with the help of binomial expansion and hence observe which term will independent of x. It is a longer process and will take time. Use the general term formula and put the power of ‘x’ in that term to 0 and hence get the value of the term independent of ‘x’. Use the surds property very carefully. One may go wrong while solving the general term of the expansion. So, be careful and don’t confuse with that step as well. One may need not calculate the value of 10! 6! And 4! In the expansion$\dfrac{10!}{4!6!}$. As we can write 10! As $10\times 9\times 8\times 7\times 6!$, so, replace 10! By this term and solve the expression further. Don’t calculate the whole multiplication as it will take time.