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.