If the coefficients of the r^{text{th}},(r+1) | vedclass

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(C) The general term of $(1+x)^n$ is $T_{k+1} = {^nC_k} x^k$. The coefficient of the $r^{	ext{th}}$ term is ${^nC_{r-1}}$. The coefficient of the $(r

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$14$

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(C) The general term of $(1+x)^n$ is $T_{k+1} = {^nC_k} x^k$. The coefficient of the $r^{	ext{th}}$ term is ${^nC_{r-1}}$. The coefficient of the $(r+1)^{	ext{th}}$ term is ${^nC_r}$. The coefficient of the $(r+2)^{	ext{th}}$ term is ${^nC_{r+1}}$. Given the ratio ${^nC_{r-1}} : {^nC_r} : {^nC_{r+1}} = 4 : 15 : 42$. From $\frac{{^nC_{r-1}}}{{^nC_r}} = \frac{4}{15}$,we get $\frac{r}{n-r+1} = \frac{4}{15}$ $\Rightarrow 15r = 4n - 4r + 4$ $\Rightarrow 19r - 4n = 4$ $(i)$. From $\frac{{^nC_r}}{{^nC_{r+1}}} = \frac{15}{42} = \frac{5}{14}$,we get $\frac{r+1}{n-r} = \frac{5}{14}$ $\Rightarrow 14r + 14 = 5n - 5r$ $\Rightarrow 19r - 5n = -14$ (ii). Subtracting (ii) from $(i)$: $(19r - 4n) - (19r - 5n) = 4 - (-14) \Rightarrow n = 18$. Substituting $n=18$ into $(i)$: $19r - 4(18) = 4$ $\Rightarrow 19r = 76$ $\Rightarrow r = 4$. Therefore,$n - r = 18 - 4 = 14$.

If the coefficients of the $r^{\text{th}}$,$(r+1)^{\text{th}}$,and $(r+2)^{\text{th}}$ terms in the expansion of $(1+x)^n$ are in the ratio $4:15:42$,then $n-r=$

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