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

Prove that the coefficient of $x^n$ in $(1+x)^{2n}$ is twice the coefficient of $x^n$ in $(1+x)^{2n-1}$

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  • General term in the expansion of $(1+x)^{2n}$ is $T_{r+1}=C(2n,n)x^n$
General term in the expansion of $(1+x)^{2n}$ is
$T_{r+1}=C(2n,n)x^n$
Putting $r=n$ we have
$T_{n+1}=C(2n,n)x^n$
Coefficient of $x^n=C(2n,n)$
Again general term in the expansion of $(1+x)^{2n-1}$ is
$T_{r+1}=C(2n-1,r)x^r$
Putting $r=n$ we have
$T_{n+1}=C(2n-1,n)x^n$
Coefficient of $x^n$ in the expansion of $x^n$ is $C(2n-1,n)$
According to the problem we have to prove that
$C(2n,n)=2\times C(2n-1,n)$
Or $\large\frac{2n!}{n!(2n-n)!}=$$2\times \large\frac{(2n-1)!}{n!(2n-1-n)!}$
$\large\frac{2n!}{n!n!}=$$2\times \large\frac{(2n-1)!}{n!(n-1)!}$
Multiplying $N^r$ and $D^r$ by n on RHS we have
$\large\frac{2n(2n-1)!}{n!n(n-1)!}$
i.e $\large\frac{2n!}{n!n!}=\frac{2n!}{n!n!}$ which is true.
Hence proved.
answered Jun 23, 2014 by sreemathi.v
 

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