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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

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

Vedclass · Answer

(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

Similar Questions

If $n$ is a positive integer and $f(n)$ is the coefficient of $x^n$ in the expansion of $(1+x)(1-x)^n$,then $f(2023) = $

Find $a$ if the $17^{\text{th}}$ and $18^{\text{th}}$ terms of the expansion $(2 + a)^{50}$ are equal.

Find the $r^{\text{th}}$ term from the end in the expansion of $(x+a)^{n}$.

The term independent of $x$ in the expansion of ${\left( {\frac{1}{2}{x^{1/3}} + {x^{ - 1/5}}} \right)^8}$ is:

The expansion of $(a+x)^n$ contains $15$ terms. When $x=1$,the ratio of the neighboring terms to the middle term in this expansion is $16$. Then the positive integral value of '$a$' is

Vedclass Test Series

Exam Paper Generator

Online Exam Module