If the coefficients of the r-th,(r+1)-th,and | vedclass

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(B) The coefficients of the $r$-th,$(r+1)$-th,and $(r+2)$-th terms in the expansion of $(1+x)^n$ are $^nC_{r-1}$,$^nC_r$,and $^nC_{r+1}$ respectively.

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$(3, 8)$

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(B) The coefficients of the $r$-th,$(r+1)$-th,and $(r+2)$-th terms in the expansion of $(1+x)^n$ are $^nC_{r-1}$,$^nC_r$,and $^nC_{r+1}$ respectively. It is given that,$^nC_{r-1} : ^nC_r : ^nC_{r+1} = 2 : 4 : 5$. From the first ratio: $\frac{^nC_{r-1}}{^nC_r} = \frac{2}{4} = \frac{1}{2}$ $\Rightarrow \frac{r}{n-r+1} = \frac{1}{2}$ $\Rightarrow 2r = n-r+1$ $\Rightarrow n = 3r-1$ $(i)$ From the second ratio: $\frac{^nC_r}{^nC_{r+1}} = \frac{4}{5}$ $\Rightarrow \frac{r+1}{n-r} = \frac{4}{5}$ $\Rightarrow 5r+5 = 4n-4r$ $\Rightarrow 4n = 9r+5$ (ii) Substituting $n = 3r-1$ into (ii): $4(3r-1) = 9r+5$ $12r-4 = 9r+5$ $3r = 9 \Rightarrow r = 3$. Substituting $r=3$ into $(i)$: $n = 3(3)-1 = 8$. Thus,$(r, n) = (3, 8)$.

If the coefficients of the $r$-th,$(r+1)$-th,and $(r+2)$-th terms in the expansion of $(1+x)^n$ are respectively in the ratio $2:4:5$,then $(r, n) =$

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