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If there is a term containing $x^{2r}$ in $(x+\large\frac{1}{x^2})^{n-3}$ then

$\begin{array}{1 1}(A)\;\text{n-2r is a positive integral multiple of 3}\\(B)\;\text{n-2r is even}\\(C)\;\text{n-2r is odd}\\(D)\;\text{none of these}\end{array} $

1 Answer

  • $T_{r+1}=nC_r a^{n-r} b^r$
$T_{k+1}$ in $(x+\large\frac{1}{x^2})^{n-3}$
$\Rightarrow n-3 C_k x^{n-3-k} (\large\frac{1}{x^2})^k$
$\Rightarrow n-3 C_k x^{n-3-3k}$
Let $T_{r+1}$ contains $x^{2r}$
$\therefore n-3-3^k=2r$
$\therefore n-2r$ is a positive integral multiple of 3
Hence (A) is the correct answer.
answered Jun 25, 2014 by sreemathi.v

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