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Home  >>  CBSE XI  >>  Math  >>  Binomial Theorem

Find the coefficient of $x^{15}$ in the expression of $(x-x^2)^{10}$

$\begin{array}{1 1}(A)\;252\\(B)\;-252\\(C)\;255\\(D)\;-255\end{array} $

1 Answer

Toolbox:
  • The general term in the expansion of $(a-b)^n$
  • $T_{r+1}=(-1)^rnC_r a^{n-r}b^r$
$(x-x^2)^{10}$
$ T_{r+1}=(-1)^r 10C_r (x)^{10-r}(x^2)^r$
$\Rightarrow (-1)^r10C_r x^{10-r}x^{2r}$
$\Rightarrow (-1)^r10C_r x^{10-r+2r}$
$\Rightarrow $ Now for this is to contain $x^{15}$ we observe that
$10+r=15$
$r=15-10$
$r=5$
Thus the coefficient of $x^{15}$ is
$(-1)^510C_5=(-1)^5.\large\frac{10!}{5!5!}$
$\Rightarrow (-1) \large\frac{10\times 9\times 8\times 7\times 5!}{5\times 4\times 3\times 2\times 5!}$
$\Rightarrow -252$
Hence (B) is the correct answer.
answered Jun 24, 2014 by sreemathi.v
 

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