Question

Find the term independent of x in the expansion: \[{{\left( 2{{x}^{2}}-\dfrac{3}{{{x}^{3}}} \right)}^{25}}\]
(a) \[\dfrac{25!}{10!15!}{{2}^{15}}{{3}^{-10}}\]
(b) \[\dfrac{25!}{10!5!}{{2}^{10}}{{3}^{-15}}\]
(c) \[\dfrac{25!}{10!15!}{{2}^{15}}{{3}^{10}}\]
(d) \[\dfrac{25!}{10!5!}{{2}^{15}}{{3}^{-10}}\]

Answer 1

Hint: Assume that the term independent of x in the expansion of \[{{\left( 2{{x}^{2}}-\dfrac{3}{{{x}^{3}}} \right)}^{25}}\] is \[{{r}^{th}}\] term. Write the expression for \[{{r}^{th}}\] term using Binomial Theorem and equate the powers of x to zero. Simplify the equation to find the value of r and thus, the value of the term.

Complete step-by-step answer:
We have to find the term independent of x in the expansion of \[{{\left( 2{{x}^{2}}-\dfrac{3}{{{x}^{3}}} \right)}^{25}}\]. We know that the term independent of x has the power of x equal to zero.
Let’s assume that the term independent of x is the \[{{r}^{th}}\] term.
We know that \[{{k}^{th}}\] term in the expansion of \[{{\left( a+b \right)}^{n}}\] is given by \[{}^{n}{{C}_{k}}{{a}^{k}}{{b}^{n-k}}\].
Substituting \[k=r,a=2{{x}^{2}},b=\dfrac{-3}{{{x}^{3}}},n=25\] in the above expression, the \[{{r}^{th}}\] term of \[{{\left( 2{{x}^{2}}-\dfrac{3}{{{x}^{3}}} \right)}^{25}}\] is given by \[{}^{25}{{C}_{r}}{{\left( 2{{x}^{2}} \right)}^{r}}{{\left( \dfrac{-3}{{{x}^{3}}} \right)}^{25-r}}\].

We can rearrange this term and write it as \[{}^{25}{{C}_{r}}{{\left( 2{{x}^{2}} \right)}^{r}}{{\left( \dfrac{-3}{{{x}^{3}}} \right)}^{25-r}}={}^{25}{{C}_{r}}{{\left( 2 \right)}^{r}}{{\left( -3 \right)}^{25-r}}{{x}^{2r-3\left( 25-r \right)}}\].
Thus, in the \[{{r}^{th}}\] term, the power of x is given by \[2r-3\left( 25-r \right)\].
We know that \[{{r}^{th}}\] term is independent of x. Thus the power of x in this term is equal to zero.
So, we have \[2r-3\left( 25-r \right)=0\].
Simplifying the above expression, we have \[2r-75+3r=0\].
Rearranging the terms of the above equation, we have \[5r=75\].
Thus, we have \[r=\dfrac{75}{5}=15\].
So, the \[{{r}^{th}}={{15}^{th}}\] term of the expansion of \[{{\left( 2{{x}^{2}}-\dfrac{3}{{{x}^{3}}} \right)}^{25}}\] is given by \[{}^{25}{{C}_{r}}{{\left( 2 \right)}^{r}}{{\left( -3 \right)}^{25-r}}{{x}^{2r-3\left( 25-r \right)}}={}^{25}{{C}_{15}}{{2}^{15}}{{\left( -3 \right)}^{25-15}}{{x}^{0}}\].
Simplifying the above expression, we have the \[{{15}^{th}}\] term as \[{}^{25}{{C}_{15}}{{2}^{15}}{{\left( -3 \right)}^{25-15}}=\dfrac{25!}{15!10!}{{2}^{15}}{{\left( -3 \right)}^{10}}=\dfrac{25!}{15!10!}{{2}^{15}}{{3}^{10}}\].
Hence, the term independent of x in the expansion of \[{{\left( 2{{x}^{2}}-\dfrac{3}{{{x}^{3}}} \right)}^{25}}\] is given by \[\dfrac{25!}{15!10!}{{2}^{15}}{{3}^{10}}\], which is option (c).

Note: One must be careful while writing the expansion of \[{{r}^{th}}\] term. We also need to know that \[{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\]. We can’t solve this question without using the fact that the term independent of x must have the power of x to be equal to zero.