If three consecutive coefficients in the bino | vedclass

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(D) Let the three consecutive coefficients be $^{n}C_{r-1}, ^{n}C_{r},$ and $^{n}C_{r+1}$.Given the ratio: $\frac{^{n}C_{r-1}}{^{n}C_{r}} = \frac{2}{1

Vedclass · Accepted Answer

$232$

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(D) Let the three consecutive coefficients be $^{n}C_{r-1}, ^{n}C_{r},$ and $^{n}C_{r+1}$.Given the ratio: $\frac{^{n}C_{r-1}}{^{n}C_{r}} = \frac{2}{15}$ and $\frac{^{n}C_{r}}{^{n}C_{r+1}} = \frac{15}{70} = \frac{3}{14}$.Using the formula $\frac{^{n}C_{r-1}}{^{n}C_{r}} = \frac{r}{n-r+1}$,we get $\frac{r}{n-r+1} = \frac{2}{15}$ $\Rightarrow 15r = 2n - 2r + 2$ $\Rightarrow 2n - 17r = -2 \dots (1)$.Using the formula $\frac{^{n}C_{r}}{^{n}C_{r+1}} = \frac{r+1}{n-r}$,we get $\frac{r+1}{n-r} = \frac{3}{14}$ $\Rightarrow 14r + 14 = 3n - 3r$ $\Rightarrow 3n - 17r = 14 \dots (2)$.Subtracting $(1)$ from $(2)$: $(3n - 17r) - (2n - 17r) = 14 - (-2) \Rightarrow n = 16$.Substituting $n = 16$ in $(1)$: $2(16) - 17r = -2$ $\Rightarrow 32 + 2 = 17r$ $\Rightarrow 17r = 34$ $\Rightarrow r = 2$.The coefficients are $^{16}C_{1}, ^{16}C_{2}, ^{16}C_{3}$.$^{16}C_{1} = 16, ^{16}C_{2} = \frac{16 	imes 15}{2} = 120, ^{16}C_{3} = \frac{16 	imes 15 	imes 14}{3 	imes 2 	imes 1} = 560$.Average $= \frac{16 + 120 + 560}{3} = \frac{696}{3} = 232$.

If three consecutive coefficients in the binomial expansion of $(x + 1)^n$ in powers of $x$ are in the ratio $2 : 15 : 70$,then the average of these three coefficients is

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