If in the expansion of (1 + x)^n,a, b, c are | vedclass

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(B) Let the three consecutive coefficients be $a = {^nC_{r-1}}$,$b = {^nC_r}$,and $c = {^nC_{r+1}}$.We know that $\frac{b}{a} = \frac{n-r+1}{r}$ and $

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$\frac{2ac + ab + bc}{b^2 - ac}$

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(B) Let the three consecutive coefficients be $a = {^nC_{r-1}}$,$b = {^nC_r}$,and $c = {^nC_{r+1}}$.We know that $\frac{b}{a} = \frac{n-r+1}{r}$ and $\frac{c}{b} = \frac{n-r}{r+1}$.From $\frac{b}{a} = \frac{n-r+1}{r}$,we get $br = an - ar + a$,so $r(a+b) = a(n+1)$,which implies $r = \frac{a(n+1)}{a+b}$.From $\frac{c}{b} = \frac{n-r}{r+1}$,we get $cr + c = bn - br$,so $r(b+c) = bn - c$,which implies $r = \frac{bn-c}{b+c}$.Equating the two expressions for $r$: $\frac{a(n+1)}{a+b} = \frac{bn-c}{b+c}$.$a(n+1)(b+c) = (bn-c)(a+b)$.$n(ab + ac) + ab + ac = abn + b^2n - ac - bc$.$n(ab + ac - ab - b^2) = -ab - ac - ac - bc$.$n(ac - b^2) = -(ab + 2ac + bc)$.$n = \frac{ab + 2ac + bc}{b^2 - ac}$.

If in the expansion of $(1 + x)^n$,$a, b, c$ are three consecutive coefficients,then $n=$

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