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BITSAT 2010
Exam
BITSAT
Year
2010
Subject
Mathematics
Chapter
Sequences and Series

BITSAT 2010 · Mathematics · Sequences and Series

The coefficient of x^ 32 in the expansion of ( (x^4-(1)/(x^3) )^ 15...

The coefficient of x32x^{32} in the expansion of (\left(x^4-\frac{1}{x^3}\right)^{15}is:is:

Options

  1. A({ }^{-15} C3\mathrm{C}_{3})
  2. B({ }^{15} C4\mathrm{C}_{4})
  3. C({ }^{-15} C5\mathrm{C}_{5})
  4. D({ }^{15} C2\mathrm{C}_{2})

Correct answer

B. ({ }^{15} C4\mathrm{C}_{4})

Step-by-step 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} \mathrmx^{\mathrm{n}-1} \cdot \mathrm{a}+{ }^{\mathrm{n}} \mathrm{C}}_{2} \mathrmx^{ \mathrm{n}-2} \mathrm{a}^2 \\ &+{ }^{\mathrm{n}} \mathrm{C}}_{3} \mathrmx^{\mathrm{n}-3} \mathrm{a}^3+{ }^{\mathrm{n}} \mathrm{C}}_{4} \mathrmx^{ \mathrm{n}-4} \cdot \mathrm{a}^4+\ldots..+{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}} \mathrm{x}^0 a^{\mathrm{n}} \end{aligned})Givenexpansionis((x4−1x3)15)Oncomparingweget(n=15,) Given expansion is (\left(x⁴-\frac{1}{x^3}\right)^{15}) On comparing we get (\mathrm{n}=15,\mathrm{x}=\mathrm{x}4),(a=(−1x3))(^4), (a=\left(-\frac{1}{x^3}\right)) (\begin{aligned} & \therefore\left(x⁴-\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}} \mathrmx^{60-7 \mathrm{r}}\\ & \Rightarrow \mathrm{x}{ }^{60-7 \mathrm{r}}=\mathrmx^{32} \Rightarrow60-7 \mathrm{r}=32 \\ & \Rightarrow 7 \mathrm{r}=28 \Rightarrow \mathrm{r}=4 \end{aligned}) So, 5 th term, contains (\mathrmx^{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} \mathrmx^{32}.) Thus, coefficient of (\mathrmx^{32}={ }^{15} \mathrm{C}}_{4}).
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