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Home  >>  CBSE XI  >>  Math  >>  Binomial Theorem
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Find the coefficient of $\large\frac{1}{x^{17}}$ in the expansion of $(x^4-\large\frac{1}{x^3})^{15}$

$\begin{array}{1 1}(A)\;1265\\(B)\;-1265\\(C)\;1365\\(D)\;-1365\end{array} $

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1 Answer

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  • The general term in the expansion of $(a-b)^n$ is $T_{r+1}=(-1)^rnC_r a^{n-r} b^r$
$T_{r+1}=(-1)^r15C_r (x^4)^{15-r} (\large\frac{1}{x^3})^r$
$\Rightarrow (-1)^r 15C_r x^{60-4r}.\large\frac{1}{x^{3r}}$
$\Rightarrow (-1)^r 15C_r x^{60-4r-3r}$
Now for this is to contain $\large\frac{1}{x^{17}}$ we observe that
$T_{r+1}=(-1)^{11} 15C_{11}$
$\Rightarrow (-1)^{11} \large\frac{15!}{11!4!}$
$\Rightarrow (-1)^{11} \large\frac{15\times 14\times 13\times 12\times 11}{11!4\times 3\times 2}$
$\Rightarrow -1365$
Hence (D) is the correct answer.
answered Jun 24, 2014 by sreemathi.v

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