Question

Find the term independent of $x$in the expansion of
${\left( {{x^3} - \dfrac{3}{{{x^2}}}} \right)^{15}}$

Answer 1

Hint: Use binomial expansion and equate the power of x to zero.

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 $b = {x^3},{\text{ }}a = \dfrac{3}{{{x^2}}},{\text{ }}n = 15$
$\Rightarrow {\left( {{x^3} - \dfrac{3}{{{x^2}}}} \right)^{15}} = \sum\limits_{r = 0}^{15} {{}^{15}{C_r}{{\left( {{x^3}} \right)}^{15 - r}}{{\left( { - \dfrac{3}{{{x^2}}}} \right)}^r}} \\\ = \sum\limits_{r = 0}^{15} {{}^{15}{C_r}{{\left( x \right)}^{45 - 3r}}{{\left( { - 1} \right)}^r}{{\left( 3 \right)}^r}{{\left( x \right)}^{ - 2r}}} = \sum\limits_{r = 0}^{15} {{}^{15}{C_r}{{\left( x \right)}^{45 - 5r}}{{\left( { - 1} \right)}^r}{{\left( 3 \right)}^r}} \\\$
Now, we want the term independent of $x$
So, put the power of $x$in the expansion of ${\left( {{x^3} - \dfrac{3}{{{x^2}}}} \right)^{15}}$ equal to zero.
$\Rightarrow 45 - 5r = 0 \\\ \Rightarrow 5r = 45 \\\ \Rightarrow r = 9 \\\$
So, put $r = 9,$in $\sum\limits_{r = 0}^{15} {{}^{15}{C_r}{{\left( x \right)}^{45 - 5r}}{{\left( { - 1} \right)}^r}{{\left( 3 \right)}^r}}$ we have
$\Rightarrow \sum\limits_{r = 0}^{15} {{}^{15}{C_r}{{\left( x \right)}^{45 - 5r}}{{\left( { - 1} \right)}^r}{{\left( 3 \right)}^r}} = {}^{15}{C_9}{\left( x \right)^0}{\left( { - 1} \right)^9}{\left( 3 \right)^9} \\\ \Rightarrow - {}^{15}{C_9}{\left( 3 \right)^9} \\\$
So, this is the required term independent of $x$ in the expansion of ${\left( {{x^3} - \dfrac{3}{{{x^2}}}} \right)^{15}}$.

Note: - Whenever we face such type of problem the key concept we have to remember is that always remember the general expansion of ${\left( {b - a} \right)^n}$, then in the expansion put the power of $x$ equal to zero, and calculate the value of $r$, then put this value of $r$ in the expansion we will get the required term which is independent of $x$.