Question

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

Answer 1

Hint:Here, first we have to expand ${{\left( {{x}^{3}}-\dfrac{1}{{{x}^{2}}} \right)}^{n}},n\in N$ with the help of binomial theorem, and then write the terms of ${{x}^{5}}$ and ${{x}^{10}}$, and find two equations for r and r’. Then, equate the sum of the coefficients of ${{x}^{5}}$ and ${{x}^{10}}$ to zero and obtain an equation for n in terms of r and r’. From the equations of r and r’ we will get the value of n.

Complete step-by-step answer:
Here, we are given ${{\left( {{x}^{3}}-\dfrac{1}{{{x}^{2}}} \right)}^{n}},n\in N$ and also given that the sum of the coefficient of ${{x}^{5}}$ and ${{x}^{10}}$ is 0.
Now, we have to find the value of n.
We know by binomial theorem that:
${{(a+b)}^{n}}={{a}^{n}}+{}^{n}{{C}_{1}}{{a}^{n-1}}b+{}^{n}{{C}_{2}}{{a}^{n-2}}{{b}^{2}}+.....+{}^{n}{{C}_{n-1}}a{{b}^{n-1}}+{{b}^{n}}$
The general term can be written as,
${}^{n}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}$
Now, consider ${{\left( {{x}^{3}}-\dfrac{1}{{{x}^{2}}} \right)}^{n}},n\in N$, by binomial expansion we get:
${{\left( {{x}^{3}}-\dfrac{1}{{{x}^{2}}} \right)}^{n}}={{\left( {{x}^{3}} \right)}^{n}}+{}^{n}{{C}_{1}}{{\left( {{x}^{3}} \right)}^{n}}\left( \dfrac{1}{{{x}^{2}}} \right)+......+{}^{n}{{C}_{n-1}}{{x}^{3}}{{\left( \dfrac{1}{{{x}^{2}}} \right)}^{n-1}}+{{\left( \dfrac{1}{{{x}^{2}}} \right)}^{n}}$
Now, let us find the term of ${{x}^{5}}$, we can take,
Term of ${{x}^{5}}={}^{n}{{C}_{r}}{{x}^{3r}}{{\left( -\dfrac{1}{{{x}^{2}}} \right)}^{n-r}}$
We know that ${{\left( \dfrac{a}{b} \right)}^{m}}=\dfrac{{{a}^{m}}}{{{b}^{m}}}$
Term of ${{x}^{5}}={}^{n}{{C}_{r}}{{x}^{3r}}{{\dfrac{(-1)}{{{\left( {{x}^{2}} \right)}^{n-r}}}}^{n-r}}$
We also know that ${{({{a}^{m}})}^{n}}={{a}^{mn}}$
Hence, we can write:
Term of ${{x}^{5}}={}^{n}{{C}_{r}}{{x}^{3r}}{{\dfrac{(-1)}{{{x}^{2(}}^{n-r)}}}^{n-r}}$
$\Rightarrow $ Term of ${{x}^{5}}={}^{n}{{C}_{r}}{{x}^{3r}}{{\dfrac{(-1)}{{{x}^{2n-2r}}}}^{n-r}}$
We also have that $\dfrac{1}{{{a}^{m}}}={{a}^{-m}}$.
Therefore, we can say that:
Term of ${{x}^{5}}={}^{n}{{C}_{r}}{{x}^{3r}}{{x}^{-(2n-2r)}}{{(-1)}^{n-r}}$
$\Rightarrow $ Term of ${{x}^{5}}={}^{n}{{C}_{r}}{{x}^{3r}}{{x}^{2r-2n}}{{(-1)}^{n-r}}$
By applying ${{x}^{m}}{{x}^{n}}={{x}^{m+n}}$,
$\Rightarrow $ Term of ${{x}^{5}}={}^{n}{{C}_{r}}{{x}^{3r+}}^{2r-2n}{{(-1)}^{n-r}}$
Now, by equating the power of x on both the sides we get:
$\begin{align}
  & 5=3r+2r-2n \\
 & \Rightarrow 5=5r-2n \\
\end{align}$
Now, by taking -2n to the left side, we get:
$5+2n=5r$
Next by cross multiplication, we obtain:
$r=\dfrac{5+2n}{5}$ …… (1)
Now, let us consider ${{x}^{10}}$ be ${{(r'+1)}^{th}}$ term. Then we can say that,
Term of ${{x}^{10}}={}^{n}{{C}_{r'}}{{x}^{3r'}}{{\left( -\dfrac{1}{{{x}^{2}}} \right)}^{n-r'}}$
Now, again by the same procedure, above, we will get:
$\Rightarrow $ Term of ${{x}^{10}}={}^{n}{{C}_{r'}}{{x}^{3r'+}}^{2r'-2n}{{(-1)}^{n-r'}}$
Hence, by equating the power of x on both the sides we get:
$\begin{align}
  & 10=3r'+2r'-2n \\
 & \Rightarrow 10=5r'-2n \\
\end{align}$
Now, by taking -2n to the left side, we get:
$10+2n=5r'$
Next by cross multiplication, we obtain:
$r'=\dfrac{10+2n}{5}$ …… (2)
We are given that the sum of the coefficients of ${{x}^{5}}$ and ${{x}^{10}}$ is zero.
Therefore, let us add the coefficients of ${{x}^{5}}$ and ${{x}^{10}}$, we obtain:
${}^{n}{{C}_{r}}{{(-1)}^{n-r}}+{}^{n}{{C}_{r'}}{{(-1)}^{n-r'}}=0$
Next, by taking ${}^{n}{{C}_{r'}}{{(-1)}^{n-r'}}$ to the right side we get:
${}^{n}{{C}_{r}}{{(-1)}^{n-r}}=-{}^{n}{{C}_{r'}}{{(-1)}^{n-r'}}$
By taking modulus on both the sides we get:
$\begin{align}
  & \left| {}^{n}{{C}_{r}}{{(-1)}^{n-r}} \right|=\left| -{}^{n}{{C}_{r'}}{{(-1)}^{n-r'}} \right| \\
 & \Rightarrow \left| {}^{n}{{C}_{r}} \right|=\left| {}^{n}{{C}_{r'}} \right| \\
 & \Rightarrow {}^{n}{{C}_{r}}={}^{n}{{C}_{r'}} \\
\end{align}$
The above condition is possible only if \[r=r'\]or \[r=n-r'\].
But r = r’ is not possible. Therefore, we can say that,
\[\begin{align}
  & r=n-r' \\
 & \Rightarrow n=r+r' \\
\end{align}\]
Now, substitute equation (1) and equation (2) in place of r and r’,
$\Rightarrow n=\dfrac{5+2n}{5}+\dfrac{10+2n}{5}$
Next, by taking LCM,
$\begin{align}
  & \Rightarrow n=\dfrac{5+2n+10+2n}{5} \\
 & \Rightarrow n=\dfrac{4n+15}{5} \\
\end{align}$
Next, by cross multiplication,
$\Rightarrow 5n=4n+15$
Now, by taking 4n to the left side,
$\begin{align}
  & \Rightarrow 5n-4n=15 \\
 & \Rightarrow n=15 \\
\end{align}$
Hence, we can say that the value of n in the expansion of ${{\left( {{x}^{3}}-\dfrac{1}{{{x}^{2}}} \right)}^{n}},n\in N$ when the sum of the coefficient of ${{x}^{5}}$ and ${{x}^{10}}$ is 0 is n = 15.
Therefore, the correct answer for this question is option (c).

Note: Here, to avoid the negative sign in the equation $-{}^{n}{{C}_{r'}}{{(-1)}^{n-r'}}$ we are taking modulus on both the sides. While taking the modulus the negative sign will be removed and the value of ${{(-1)}^{n-r'}}$ and ${{(-1)}^{n-r}}$ becomes positive 1.