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

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

Vedclass · Accepted Answer

$n^2 - n(4p + 1) + 4p^2 - 2 = 0$

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(B) The coefficients of the $p^{th}$,$(p + 1)^{th}$,and $(p + 2)^{th}$ terms in the expansion of $(1 + x)^n$ are $^nC_{p-1}$,$^nC_p$,and $^nC_{p+1}$.Since they are in $A.P.$,we have $2(^nC_p) = ^nC_{p-1} + ^nC_{p+1}$.Using the formula $^nC_r = \frac{n!}{r!(n-r)!}$,we get:$2 \frac{n!}{p!(n-p)!} = \frac{n!}{(p-1)!(n-p+1)!} + \frac{n!}{(p+1)!(n-p-1)!}$.Dividing by $n!$ and multiplying by $p!(n-p+1)!$,we simplify the expression to:$2(n-p+1) = p(n-p+1) + p(p+1)$ is incorrect; the standard simplification leads to:$n^2 - n(4p + 1) + 4p^2 - 2 = 0$.Verification: Let $p = 1$. Then $^nC_0, ^nC_1, ^nC_2$ are in $A.P.$$2(^nC_1) = ^nC_0 + ^nC_2$ $\Rightarrow 2n = 1 + \frac{n(n-1)}{2}$ $\Rightarrow 4n = 2 + n^2 - n$ $\Rightarrow n^2 - 5n + 2 = 0$.Substituting $p=1$ into option $(b)$: $n^2 - n(4(1) + 1) + 4(1)^2 - 2 = n^2 - 5n + 2 = 0$. This matches.

If the coefficients of the $p^{th}$,$(p + 1)^{th}$,and $(p + 2)^{th}$ terms in the expansion of $(1 + x)^n$ are in $A.P.$,then

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