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Home  >>  CBSE XI  >>  Math  >>  Binomial Theorem
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If $x^p$ occurs in the expansion of $(x^2+\large\frac{1}{x})^{2n}$.Prove that its coefficient is $\large\frac{2n!}{\Large\frac{4n-p}{3}!\frac{2n+p}{3}!}$

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Toolbox:
  • $T_{r+1}=nC_r(a)^{n-r}b^r$
In the expansion of $(x^2+\large\frac{1}{x})^{2n}$ we have
$t_{r+1}=2nC_r.(x^2)^{2n-r}.(\large\frac{1}{x})^r$
$\Rightarrow 2nC_r .x^{4n-2r} .\large\frac{1}{x^r}$
$\Rightarrow 2nC_r .x^{4n-3r} $
Putting $(4n-3r)=p$ we get
$r=\large\frac{1}{3}$$(4n-p)$
Step 2:
Coefficient of $x^p=2nC_{(1/3)(4n-p)}$
$\Rightarrow \large\frac{2n!}{[\Large\frac{1}{3}(4n-p)!]\times [2n-\Large\frac{1}{3}(4n-p)]!}$
$\Rightarrow \large\frac{2n!}{[\Large\frac{4n-p}{3}]![\Large\frac{2n+p}{3}]!}$
Hence proved.
answered Jun 24, 2014 by sreemathi.v
 

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