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(C) Let the three consecutive coefficients of $(1+x)^n$ be $^nC_{r-1}, ^nC_r, ^nC_{r+1}$.By the given condition,$^nC_{r-1} : ^nC_r : ^nC_{r+1} = 6 : 3

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$12$

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(C) Let the three consecutive coefficients of $(1+x)^n$ be $^nC_{r-1}, ^nC_r, ^nC_{r+1}$.By the given condition,$^nC_{r-1} : ^nC_r : ^nC_{r+1} = 6 : 33 : 110$.Using the ratio of consecutive binomial coefficients $\frac{^nC_r}{^nC_{r-1}} = \frac{n-r+1}{r}$,we have:$\frac{^nC_r}{^nC_{r-1}} = \frac{33}{6} = \frac{11}{2}$ $\Rightarrow 2(n-r+1) = 11r$ $\Rightarrow 2n - 13r + 2 = 0$ $(i)$.Similarly,$\frac{^nC_{r+1}}{^nC_r} = \frac{110}{33} = \frac{10}{3}$ $\Rightarrow 3(n-r) = 10(r+1)$ $\Rightarrow 3n - 13r - 10 = 0$ (ii).Subtracting $(i)$ from (ii),we get $(3n - 13r - 10) - (2n - 13r + 2) = 0$ $\Rightarrow n - 12 = 0$ $\Rightarrow n = 12$.Substituting $n=12$ in $(i)$,$24 - 13r + 2 = 0$ $\Rightarrow 13r = 26$ $\Rightarrow r = 2$.

If $n$ is a positive integer and three consecutive coefficients in the expansion of $(1 + x)^n$ are in the ratio $6 : 33 : 110$,then $n =$

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