If A and B are coefficients of {x^n} in expansions of {(1 + x)^{2n}} and {(1 + x)^{2n

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7.Binomial Theorem

medium

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

$A = B$

$A = 2B$

$2A = B$

None of these

Solution

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

$ = \frac{{^{2n}{C_n}}}{{^{(2n – 1)}{C_n}}} = \frac{{(2n)!}}{{n!\,n!}} \times \frac{{(n – 1)!\,n!}}{{(2n – 1)!}}$

$ = \frac{{(2n)(2n – 1)!(n – 1)!}}{{n(n – 1)!\,\,(2n – 1)!}} = \frac{{2n}}{n} = 2:1$

==> $\frac{A}{B} = \frac{2}{1}$

==> $A = 2B$.

Standard 11

Mathematics

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hard

(JEE MAIN-2023)

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

Solution

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