7.Binomial Theorem
medium

જો ${(1 + x)^{2n}}$ અને ${(1 + x)^{2n - 1}}$ ની વિસ્તરણમાં $A$ અને $B$ એ ${x^n}$ ના સહગુણક હોય તો . . . .

A

$A = B$

B

$A = 2B$

C

$2A = B$

D

એકપણ નહીં.

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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