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In the expansion of $\left(x^{3}-\frac{1}{x^{2}}\right)^{n}$ where n is a positive integer, the sum of the coefficients of $\mathrm{x}^{5}$ and $\mathrm{x}^{10}$ is $0 .$
What is the value of the independent term?
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What is the value of the independent term?
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Verified Answer
The correct answer is:
$-5005$
$\left(x^{3}-\frac{1}{x^{2}}\right)^{n}$
$\quad$ General term, $T_{r+1}={ }^{n} C_{r}\left(x^{3}\right)^{n-r} \cdot\left(-\frac{1}{x^{2}}\right)^{r}$
$\quad={ }^{n} C_{r} \cdot x^{(3 n-3 r)} \cdot(-1)^{r} \cdot x^{-2 r}$
$={ }^{n} C_{r} \cdot(-1)^{r} \cdot x^{(3 n-5 r)}$ ...(i)
For the coefficient $\mathrm{x}^{5}$
Put $3 n-5 r=5$
$5 r=3 n-5$
$\therefore \mathrm{r}=\frac{3 \mathrm{n}}{5}-1$
$\therefore$ Coefficient of $x^{5}={ }^{n} C\left(\frac{3 n}{5}-1\right)^{(-1)}\left(\frac{3 n}{5}-1\right)$
For the coefficient of $\mathrm{x}^{10}$
Put $3 n-5 r=10$
$5 r=3 n-10$
$\therefore \mathrm{r}=\frac{3 \mathrm{n}}{5}-2$
$\therefore$ Coefficient of $x^{10}={ }^{n} C_{\left(\frac{3 n}{5}-2\right)^{(-1)}}\left(\frac{3 n}{5}-2\right)$
The sum of the coefficient of $x^{5}$ and $x^{10}=0$
$\Rightarrow{ }^{\mathrm{n}} \mathrm{C}_{\left(\frac{3 \mathrm{n}}{5}-1\right)^{(-1)}\left(\frac{3 \mathrm{n}}{5}-1\right)}+{ }^{\mathrm{n}} \mathrm{C}_{\left(\frac{3 \mathrm{n}}{5}-2\right)}(-1)^{\left(\frac{3 \mathrm{n}}{5}-2\right)}=0$
$\Rightarrow$
$\left.(-1)^{\frac{3 n}{5}}\left[{ }^{n} C_{\left(\frac{3 n}{5}-1\right)} \cdot(-1)^{-1}+{ }^{n} C_{\left(\frac{3 n}{5}-2\right.}\right) \cdot(-1)^{(-2)}\right]=0$
$\Rightarrow-{ }^{n} C_{\left(\frac{3 n}{5}-1\right)}+{ }^{n} C_{\left(\frac{3 n}{5}-2\right)}=0$ ...(ii)
For the independent term, put $3 \mathrm{n}-5 \mathrm{r}=0 \quad[$
from eq. $(\mathrm{i})]$
$\Rightarrow 5 \mathrm{r}=3 \mathrm{n}=3 \times 15$
$5 \mathrm{r}=3 \times 3 \times 5$
$\mathrm{r}=9$
Putting the value of r in eq. (i), we get $\mathrm{T}_{9+1}={ }^{15} \mathrm{C}_{9} \cdot(-1)^{9} \cdot \mathrm{x}^{(3 \times 15-5 \times 9)}$
$\Rightarrow \mathrm{T}_{10}=-{ }^{15} \mathrm{C}_{9} \cdot \mathrm{x}^{0}=-{ }^{15} \mathrm{C}_{9}$
$\Rightarrow \mathrm{T}_{10}=-{ }^{15} \mathrm{C}_{6} \quad\left[\because{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}-\mathrm{r}}\right]$
$=\frac{-15 !}{6 ! 9 !}$
$\left[\because{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}=\frac{\mathrm{n} !}{\mathrm{r} !(\mathrm{n}-\mathrm{r}) !}\right]$
$=-5005$
$\quad$ General term, $T_{r+1}={ }^{n} C_{r}\left(x^{3}\right)^{n-r} \cdot\left(-\frac{1}{x^{2}}\right)^{r}$
$\quad={ }^{n} C_{r} \cdot x^{(3 n-3 r)} \cdot(-1)^{r} \cdot x^{-2 r}$
$={ }^{n} C_{r} \cdot(-1)^{r} \cdot x^{(3 n-5 r)}$ ...(i)
For the coefficient $\mathrm{x}^{5}$
Put $3 n-5 r=5$
$5 r=3 n-5$
$\therefore \mathrm{r}=\frac{3 \mathrm{n}}{5}-1$
$\therefore$ Coefficient of $x^{5}={ }^{n} C\left(\frac{3 n}{5}-1\right)^{(-1)}\left(\frac{3 n}{5}-1\right)$
For the coefficient of $\mathrm{x}^{10}$
Put $3 n-5 r=10$
$5 r=3 n-10$
$\therefore \mathrm{r}=\frac{3 \mathrm{n}}{5}-2$
$\therefore$ Coefficient of $x^{10}={ }^{n} C_{\left(\frac{3 n}{5}-2\right)^{(-1)}}\left(\frac{3 n}{5}-2\right)$
The sum of the coefficient of $x^{5}$ and $x^{10}=0$
$\Rightarrow{ }^{\mathrm{n}} \mathrm{C}_{\left(\frac{3 \mathrm{n}}{5}-1\right)^{(-1)}\left(\frac{3 \mathrm{n}}{5}-1\right)}+{ }^{\mathrm{n}} \mathrm{C}_{\left(\frac{3 \mathrm{n}}{5}-2\right)}(-1)^{\left(\frac{3 \mathrm{n}}{5}-2\right)}=0$
$\Rightarrow$
$\left.(-1)^{\frac{3 n}{5}}\left[{ }^{n} C_{\left(\frac{3 n}{5}-1\right)} \cdot(-1)^{-1}+{ }^{n} C_{\left(\frac{3 n}{5}-2\right.}\right) \cdot(-1)^{(-2)}\right]=0$
$\Rightarrow-{ }^{n} C_{\left(\frac{3 n}{5}-1\right)}+{ }^{n} C_{\left(\frac{3 n}{5}-2\right)}=0$ ...(ii)
For the independent term, put $3 \mathrm{n}-5 \mathrm{r}=0 \quad[$
from eq. $(\mathrm{i})]$
$\Rightarrow 5 \mathrm{r}=3 \mathrm{n}=3 \times 15$
$5 \mathrm{r}=3 \times 3 \times 5$
$\mathrm{r}=9$
Putting the value of r in eq. (i), we get $\mathrm{T}_{9+1}={ }^{15} \mathrm{C}_{9} \cdot(-1)^{9} \cdot \mathrm{x}^{(3 \times 15-5 \times 9)}$
$\Rightarrow \mathrm{T}_{10}=-{ }^{15} \mathrm{C}_{9} \cdot \mathrm{x}^{0}=-{ }^{15} \mathrm{C}_{9}$
$\Rightarrow \mathrm{T}_{10}=-{ }^{15} \mathrm{C}_{6} \quad\left[\because{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}-\mathrm{r}}\right]$
$=\frac{-15 !}{6 ! 9 !}$
$\left[\because{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}=\frac{\mathrm{n} !}{\mathrm{r} !(\mathrm{n}-\mathrm{r}) !}\right]$
$=-5005$
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