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Question: In the expansion of \({{\left( {{x}^{3}}-\dfrac{1}{{{x}^{2}}} \right)}^{n}},n\in N\), if the sum of ...

In the expansion of (x31x2)n,nN{{\left( {{x}^{3}}-\dfrac{1}{{{x}^{2}}} \right)}^{n}},n\in N, if the sum of the coefficient of x5{{x}^{5}} and x10{{x}^{10}} is 0, then the value of n is:
(a) 25
(b) 20
(c) 15
(d) None of these

Explanation

Solution

Hint:Here, first we have to expand (x31x2)n,nN{{\left( {{x}^{3}}-\dfrac{1}{{{x}^{2}}} \right)}^{n}},n\in N with the help of binomial theorem, and then write the terms of x5{{x}^{5}} and x10{{x}^{10}}, and find two equations for r and r’. Then, equate the sum of the coefficients of x5{{x}^{5}} and x10{{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 (x31x2)n,nN{{\left( {{x}^{3}}-\dfrac{1}{{{x}^{2}}} \right)}^{n}},n\in N and also given that the sum of the coefficient of x5{{x}^{5}} and x10{{x}^{10}} is 0.
Now, we have to find the value of n.
We know by binomial theorem that:
(a+b)n=an+nC1an1b+nC2an2b2+.....+nCn1abn1+bn{{(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,
nCranrbr{}^{n}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}
Now, consider (x31x2)n,nN{{\left( {{x}^{3}}-\dfrac{1}{{{x}^{2}}} \right)}^{n}},n\in N, by binomial expansion we get:
(x31x2)n=(x3)n+nC1(x3)n(1x2)+......+nCn1x3(1x2)n1+(1x2)n{{\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 x5{{x}^{5}}, we can take,
Term of x5=nCrx3r(1x2)nr{{x}^{5}}={}^{n}{{C}_{r}}{{x}^{3r}}{{\left( -\dfrac{1}{{{x}^{2}}} \right)}^{n-r}}
We know that (ab)m=ambm{{\left( \dfrac{a}{b} \right)}^{m}}=\dfrac{{{a}^{m}}}{{{b}^{m}}}
Term of x5=nCrx3r(1)(x2)nrnr{{x}^{5}}={}^{n}{{C}_{r}}{{x}^{3r}}{{\dfrac{(-1)}{{{\left( {{x}^{2}} \right)}^{n-r}}}}^{n-r}}
We also know that (am)n=amn{{({{a}^{m}})}^{n}}={{a}^{mn}}
Hence, we can write:
Term of x5=nCrx3r(1)x2(nr)nr{{x}^{5}}={}^{n}{{C}_{r}}{{x}^{3r}}{{\dfrac{(-1)}{{{x}^{2(}}^{n-r)}}}^{n-r}}
\Rightarrow Term of x5=nCrx3r(1)x2n2rnr{{x}^{5}}={}^{n}{{C}_{r}}{{x}^{3r}}{{\dfrac{(-1)}{{{x}^{2n-2r}}}}^{n-r}}
We also have that 1am=am\dfrac{1}{{{a}^{m}}}={{a}^{-m}}.
Therefore, we can say that:
Term of x5=nCrx3rx(2n2r)(1)nr{{x}^{5}}={}^{n}{{C}_{r}}{{x}^{3r}}{{x}^{-(2n-2r)}}{{(-1)}^{n-r}}
\Rightarrow Term of x5=nCrx3rx2r2n(1)nr{{x}^{5}}={}^{n}{{C}_{r}}{{x}^{3r}}{{x}^{2r-2n}}{{(-1)}^{n-r}}
By applying xmxn=xm+n{{x}^{m}}{{x}^{n}}={{x}^{m+n}},
\Rightarrow Term of x5=nCrx3r+2r2n(1)nr{{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:
5=3r+2r2n 5=5r2n \begin{aligned} & 5=3r+2r-2n \\\ & \Rightarrow 5=5r-2n \\\ \end{aligned}
Now, by taking -2n to the left side, we get:
5+2n=5r5+2n=5r
Next by cross multiplication, we obtain:
r=5+2n5r=\dfrac{5+2n}{5} …… (1)
Now, let us consider x10{{x}^{10}} be (r+1)th{{(r'+1)}^{th}} term. Then we can say that,
Term of x10=nCrx3r(1x2)nr{{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 x10=nCrx3r+2r2n(1)nr{{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:
10=3r+2r2n 10=5r2n \begin{aligned} & 10=3r'+2r'-2n \\\ & \Rightarrow 10=5r'-2n \\\ \end{aligned}
Now, by taking -2n to the left side, we get:
10+2n=5r10+2n=5r'
Next by cross multiplication, we obtain:
r=10+2n5r'=\dfrac{10+2n}{5} …… (2)
We are given that the sum of the coefficients of x5{{x}^{5}} and x10{{x}^{10}} is zero.
Therefore, let us add the coefficients of x5{{x}^{5}} and x10{{x}^{10}}, we obtain:
nCr(1)nr+nCr(1)nr=0{}^{n}{{C}_{r}}{{(-1)}^{n-r}}+{}^{n}{{C}_{r'}}{{(-1)}^{n-r'}}=0
Next, by taking nCr(1)nr{}^{n}{{C}_{r'}}{{(-1)}^{n-r'}} to the right side we get:
nCr(1)nr=nCr(1)nr{}^{n}{{C}_{r}}{{(-1)}^{n-r}}=-{}^{n}{{C}_{r'}}{{(-1)}^{n-r'}}
By taking modulus on both the sides we get:
nCr(1)nr=nCr(1)nr nCr=nCr nCr=nCr \begin{aligned} & \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{aligned}
The above condition is possible only if r=rr=r'or r=nrr=n-r'.
But r = r’ is not possible. Therefore, we can say that,

& r=n-r' \\\ & \Rightarrow n=r+r' \\\ \end{aligned}$$ 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, $\begin{aligned} & \Rightarrow n=\dfrac{5+2n+10+2n}{5} \\\ & \Rightarrow n=\dfrac{4n+15}{5} \\\ \end{aligned}$ Next, by cross multiplication, $\Rightarrow 5n=4n+15$ Now, by taking 4n to the left side, $\begin{aligned} & \Rightarrow 5n-4n=15 \\\ & \Rightarrow n=15 \\\ \end{aligned}$ 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.