Gujarati
7.Binomial Theorem
medium

यदि ${(a + b)^n}$ के प्रसार में $\frac{{{T_2}}}{{{T_3}}}$ व ${(a + b)^{n + 3}}$ के प्रसार में $\frac{{{T_3}}}{{{T_4}}}$ समान हैं, तब $n=$

A

$3$

B

$4$

C

$5$

D

$6$

Solution

(c) प्रश्नानुसार, $\frac{{{T_2}}}{{{T_3}}} = \frac{{^n{C_1}{a^{n – 1}}b}}{{^n{C_2}{a^{n – 2}}{b^2}}}$ ……$(i)$

$\frac{{{T_2}}}{{{T_3}}} = \frac{{^n{C_1}{a^{n – 1}}b}}{{^n{C_2}{a^{n – 2}}{b^2}}}$ ……$(ii)$

$(i)$ = $(ii)$ ==> $\frac{{2n}}{{n(n – 1)}} = \frac{{6(n + 3)(n + 2)}}{{2(n + 3)(n + 2)(n + 1)}}$

==> $2(n + 1) = 3(n – 1) \Rightarrow n = 5$.

Standard 11
Mathematics

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