Question

In the expansion of ${{\left( {{x}^{3}}-\dfrac{1}{{{x}^{2}}} \right)}^{n}}$, $n\in N$, if sum of coefficients of ${{x}^{5}}$ and ${{x}^{10}}$ is 0 then n is
a)25
b)20
c)15
d)None of these

Answer 1

Hint: In this question, we have to find the coefficients of ${{x}^{5}}$ and ${{x}^{10}}$ in the expansion of the expression${{\left( {{x}^{3}}-\dfrac{1}{{{x}^{2}}} \right)}^{n}}$,$n\in N$. As the expression in the parenthesis contains two terms, we can use binomial theorem for the expansion and obtain the coefficients for particular powers of x. By comparing it with the given values in the question, we can obtain n.

Complete step-by-step answer:
In the first step, we have to expand${{\left( {{x}^{3}}-\dfrac{1}{{{x}^{2}}} \right)}^{n}}$,$n\in N$. We are given the coefficients of ${{x}^{5}}$ and ${{x}^{10}}$. Now, we can use the binomial theorem which states that
${{\left( a+b \right)}^{n}}=\sum\limits_{r=0}^{n}{\dfrac{n!}{r!\left( n-r \right)!}{{a}^{r}}{{b}^{n-r}}}........................(1.1)$
Now, we can take $a={{x}^{3}}\text{ and }b=\dfrac{-1}{{{x}^{2}}}$ in equation(1.1) to obtain
\[{{\left( {{x}^{3}}-\dfrac{1}{{{x}^{2}}} \right)}^{n}}=\sum\limits_{r=0}^{n}{\dfrac{n!}{r!\left( n-r \right)!}{{\left( {{x}^{3}} \right)}^{r}}{{\left( \dfrac{-1}{{{x}^{2}}} \right)}^{n-r}}}=\sum\limits_{r=0}^{n}{\dfrac{{{\left( -1 \right)}^{n-r}}n!}{r!\left( n-r \right)!}{{\left( x \right)}^{3r-2(n-r)}}=}\sum\limits_{r=0}^{n}{\dfrac{{{\left( -1 \right)}^{n-r}}n!}{r!\left( n-r \right)!}{{\left( x \right)}^{5r-2n}}}............(1.2)\]
We note that in this equation, all the terms have a particular power of x. Thus the coefficient of \[{{\left( x \right)}^{5r-2n}}\text{ is }\dfrac{{{\left( -1 \right)}^{n-r}}n!}{r!\left( n-r \right)!}..............(1.3)\]
For the power of x to be 5,
$5r-2n=5\Rightarrow r=\dfrac{5+2n}{5}............(1.4)$
And for the power of x to be 10
$5r-2n=10\Rightarrow r=\dfrac{10+2n}{5}.............(1.5)$

It is given that the sum of coefficients of ${{x}^{5}}$ and ${{x}^{10}}$is 0. Using this in equation (1.3), (1.4) and (1.5) we obtain,
$\text{Coefficient of }{{x}^{5}}+\text{Coefficient of }{{x}^{10}}=0$
\[\begin{align}
  & \Rightarrow \dfrac{{{\left( -1 \right)}^{n-\left( \dfrac{5+2n}{5} \right)}}n!}{\left( \dfrac{5+2n}{5} \right)!\left( n-\left( \dfrac{5+2n}{5} \right) \right)!}+\dfrac{{{\left( -1 \right)}^{n-\left( \dfrac{10+2n}{5} \right)}}n!}{\left( \dfrac{10+2n}{5} \right)!\left( n-\left( \dfrac{10+2n}{5} \right) \right)!}=0 \\
 & \Rightarrow \dfrac{{{\left( -1 \right)}^{\dfrac{3n-5}{5}}}n!}{\left( \dfrac{5+2n}{5} \right)!\left( \dfrac{3n-5}{5} \right)!}+\dfrac{{{\left( -1 \right)}^{\dfrac{3n-10}{5}}}n!}{\left( \dfrac{10+2n}{5} \right)!\left( \dfrac{3n-10}{5} \right)!}=............0 \\
 & \Rightarrow \dfrac{{{\left( -1 \right)}^{\dfrac{3n-10}{5}+1}}n!}{\left( 1+\dfrac{2n}{5} \right)!\left( \dfrac{3n}{5}-1 \right)!}=-\dfrac{{{\left( -1 \right)}^{\dfrac{3n-10}{5}}}n!}{\left( 2+\dfrac{2n}{5} \right)!\left( \dfrac{3n}{5}-2 \right)!}................(1.6) \\
\end{align}\]
Cancelling n! from the numerator and using the fact that
$\left( 2+\dfrac{2n}{5} \right)!=\left( 2+\dfrac{2n}{5} \right)\times \left( 1+\dfrac{2n}{5} \right)!$
 $\left( \dfrac{3n}{5}-1 \right)!=\left( \dfrac{3n}{5}-1 \right)\left( \dfrac{3n}{5}-2 \right)!$
And
\[{{\left( -1 \right)}^{\dfrac{3n-10}{5}+1}}=-{{\left( -1 \right)}^{\dfrac{3n-10}{5}}}\]
in equation (1.6), we can cancel out the terms in both sides to obtain
$\left( 2+\dfrac{2n}{5} \right)=\left( \dfrac{3n}{5}-1 \right)\Rightarrow \dfrac{n}{5}=3\Rightarrow n=15$
Therefore, the condition given in the question is satisfied for n=15 and thus n=15 is the answer for this question which matches the option (c) given in the question. Thus, option (c) is the correct answer.

Note: We should be careful to cancel out the common terms in equation (1.6) using the rules for the factorial rather than expanding the expression totally as it would then become a very high degree polynomial and would be difficult to solve.