If frac{{{T_2}}}{{{T_3}}} in the expansion of | vedclass

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(c) Accordingly, $\frac{{{T_2}}}{{{T_3}}} = \frac{{^n{C_1}{a^{n - 1}}b}}{{^n{C_2}{a^{n - 2}}{b^2}}}$ ......$(i)$

$\frac{{{T_2}}}{{{T_3}}} = \frac{{

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(c) Accordingly, $\frac{{{T_2}}}{{{T_3}}} = \frac{{^n{C_1}{a^{n - 1}}b}}{{^n{C_2}{a^{n - 2}}{b^2}}}$ ......$(i)$

$\frac{{{T_2}}}{{{T_3}}} = \frac{{^n{C_1}{a^{n - 1}}b}}{{^n{C_2}{a^{n - 2}}{b^2}}}$ ......$(ii)$
$(i)$ = $(ii)$ ==> $\frac{{2n}}{{n(n - 1)}} = \frac{{6(n + 3)(n + 2)}}{{2(n + 3)(n + 2)(n + 1)}}$

==> $2(n + 1) = 3(n - 1) \Rightarrow n = 5$.

If $\frac{{{T_2}}}{{{T_3}}}$ in the expansion of ${(a + b)^n}$ and $\frac{{{T_3}}}{{{T_4}}}$ in the expansion of ${(a + b)^{n + 3}}$ are equal, then $n=$

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