If frac{{{T₂}}}{{{T₃}}} in expansion of {(a + b)^n} and frac{{{T₃}}}{{{T₄}}} in expansion of

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7.Binomial Theorem

medium

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 $n=$

$3$

$4$

$5$

$6$

Solution

$\frac{{{T_2}}}{{{T_3}}} = \frac{{^n{C_1}{a^{n – 1}}b}}{{^n{C_2}{a^{n – 2}}{b^2}}}$ ……$(ii)$
$(i)$ = $(ii)$ ==> $\frac{{2n}}{{n(n – 1)}} = \frac{{6(n + 3)(n + 2)}}{{2(n + 3)(n + 2)(n + 1)}}$

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

Standard 11

Mathematics

If $n$ is even positive integer, then the condition that the greatest term in the expansion of ${(1 + x)^n}$ may have the greatest coefficient also, is

hard

If the coefficients of ${T_r},\,{T_{r + 1}},\,{T_{r + 2}}$ terms of ${(1 + x)^{14}}$ are in $A.P.$, then $r =$

medium

If the second term of the expansion ${\left[ {{a^{\frac{1}{{13}}}}\,\, + \,\,\frac{a}{{\sqrt {{a^{ – 1}}} }}} \right]^n}$ is $14a^{5/2}$ then the value of $\frac{{^n{C_3}}}{{^n{C_2}}}$ is :

normal

The number of integral terms in the expansion of $(7^{1/3} + 11^{1/9})^{6561}$ is :-