If A and B are the coefficients of {x^n} in t | vedclass

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Question

(b) $\frac{{{\rm{Coefficient}}\,{\rm{of}}\,{x^n}\,{\rm{in}}\,{\rm{expansion}}\,\,{\rm{of}}{{(1 + x)}^{2n}}}}{{{\rm{Coefficient}}\,{\rm{of}}\,{x^n}\,{\

Vedclass · Accepted Answer

$A = 2B$

Vedclass · Answer

(b) $\frac{{{m{Coefficient}}\,{m{of}}\,{x^n}\,{m{in}}\,{m{expansion}}\,\,{m{of}}{{(1 + x)}^{2n}}}}{{{m{Coefficient}}\,{m{of}}\,{x^n}\,{m{in}}\,{m{expansion}}\,\,{m{of}}\,{{(1 + x)}^{2n - 1}}}}$

$ = \frac{{^{2n}{C_n}}}{{^{(2n - 1)}{C_n}}} = \frac{{(2n)!}}{{n!\,n!}} 	imes \frac{{(n - 1)!\,n!}}{{(2n - 1)!}}$

$ = \frac{{(2n)(2n - 1)!(n - 1)!}}{{n(n - 1)!\,\,(2n - 1)!}} = \frac{{2n}}{n} = 2:1$

==> $\frac{A}{B} = \frac{2}{1}$

==> $A = 2B$.

If $A$ and $B$ are the coefficients of ${x^n}$ in the expansions of ${(1 + x)^{2n}}$ and ${(1 + x)^{2n - 1}}$ respectively, then

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