Q48 / 125 · BITSAT 2010 · Mathematics
The coefficient of \(x^{32}\) in the expansion of \(\left(x^4-\frac{1}{x^3}\right)^{15}\) is:
- A\({ }^{-15} \mathrm{C}_3\)
- Correct answerB\({ }^{15} \mathrm{C}_4\)
- C\({ }^{-15} \mathrm{C}_5\)
- D\({ }^{15} \mathrm{C}_2\)
Solution
We know by Binomial expansion, that \((x+a)^{\mathrm{n}}\)
\(\begin{aligned}
&={ }^{\mathrm{n}} \mathrm{C}_0 x^{\mathrm{n}} \mathrm{a}^0+{ }^{\mathrm{n}} \mathrm{C}_1 \mathrm{x}^{\mathrm{n}-1} \cdot \mathrm{a}+{ }^{\mathrm{n}} \mathrm{C}_2 \mathrm{x}^{\mathrm{n}-2} \mathrm{a}^2 \\
&+{ }^{\mathrm{n}} \mathrm{C}_3 \mathrm{x}^{\mathrm{n}-3} \mathrm{a}^3+{ }^{\mathrm{n}} \mathrm{C}_4 \mathrm{x}^{\mathrm{n}-4} \cdot \mathrm{a}^4+\ldots . .+{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}} \mathrm{x}^0 a^{\mathrm{n}}
\end{aligned}\)
Given expansion is \(\left(x^4-\frac{1}{x^3}\right)^{15}\)
On comparing we get \(\mathrm{n}=15, \mathrm{x}=\mathrm{x}^4\),
\(a=\left(-\frac{1}{x^3}\right)\)
\(\begin{aligned}
& \therefore\left(x^4-\frac{1}{x^3}\right)^{15}={ }^{15} \mathrm{C}_0\left(\mathrm{x}^4\right)^{15}\left(-\frac{1}{\mathrm{x}^3}\right)^0 \\
& +{ }^{15} \mathrm{C}_1\left(\mathrm{x}^4\right)^{14}\left(-\frac{1}{\mathrm{x}^3}\right)+{ }^{15} \mathrm{C}_2\left(\mathrm{x}^4\right)^{13}\left(-\frac{1}{\mathrm{x}^3}\right)^2 \\
& +{ }^{15} \mathrm{C}_3\left(\mathrm{x}^4\right)^{12}\left(-\frac{1}{\mathrm{x}^3}\right)^3+{ }^{15} \mathrm{C}_4\left(\mathrm{x}^4\right)^{11}\left(-\frac{1}{\mathrm{x}^3}\right)^4+\ldots \\
& \mathrm{T}_{\mathrm{r}+1}={ }^{15} \mathrm{C}_{\mathrm{r}}\left(\mathrm{x}^4\right)^{15-\mathrm{r}} \cdot\left(-\frac{1}{\mathrm{x}^3}\right)^{\mathrm{r}}=-{ }^{15} \mathrm{C}_{\mathrm{r}} \mathrm{x}^{60-7 \mathrm{r}} \\
& \Rightarrow \mathrm{x}{ }^{60-7 \mathrm{r}}=\mathrm{x}^{32} \Rightarrow 60-7 \mathrm{r}=32 \\
& \Rightarrow 7 \mathrm{r}=28 \Rightarrow \mathrm{r}=4
\end{aligned}\)
So, 5 th term, contains \(\mathrm{x}^{32}\)
\(={ }^{15} C_4\left(x^4\right){ }^{11}\left(-\frac{1}{x^3}\right)^4={ }^{15} C_4 x^{44} x^{-12}\)
\(={ }^{15} \mathrm{C}_4 \mathrm{x}^{32} .\)
Thus, coefficient of \(\mathrm{x}^{32}={ }^{15} \mathrm{C}_4\).
Hence, the correct answer is option: 'B'.