The sum of all possible values of n in N such | vedclass

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

Question

(D) The expansion is $(1+2x^2+x^4)(^nC_0 + ^nC_1x + ^nC_2x^2 + ^nC_3x^3 + \dots)$.The coefficient of $x$ is $^nC_1 = n$.The coefficient of $x^2$ is $2

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(D) The expansion is $(1+2x^2+x^4)(^nC_0 + ^nC_1x + ^nC_2x^2 + ^nC_3x^3 + \dots)$.The coefficient of $x$ is $^nC_1 = n$.The coefficient of $x^2$ is $2 + ^nC_2 = 2 + \frac{n(n-1)}{2}$.The coefficient of $x^3$ is $2(^nC_1) + ^nC_3 = 2n + \frac{n(n-1)(n-2)}{6}$.Since these are in arithmetic progression, $2 	imes (	ext{coeff. of } x^2) = (	ext{coeff. of } x) + (	ext{coeff. of } x^3)$.$2 \left[ 2 + \frac{n(n-1)}{2} ight] = n + 2n + \frac{n(n-1)(n-2)}{6}$.$4 + n(n-1) = 3n + \frac{n(n-1)(n-2)}{6}$.$24 + 6(n^2-n) = 18n + n(n^2-3n+2)$.$24 + 6n^2 - 6n = 18n + n^3 - 3n^2 + 2n$.$n^3 - 9n^2 + 26n - 24 = 0$.Testing values: for $n=2$, $8-36+52-24=0$ (True).For $n=3$, $27-81+78-24=0$ (True).For $n=4$, $64-144+104-24=0$ (True).The possible values of $n$ are $2, 3, 4$.The sum of these values is $2+3+4 = 9$.

The sum of all possible values of $n \in N$ such that the coefficients of $x$, $x^2$, and $x^3$ in the expansion of $(1+x^2)^2(1+x)^n$ are in arithmetic progression is:

Similar Questions

The term independent of $x$ in the expansion of ${\left( {{x^2} - \frac{1}{x}} \right)^9}$ is

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 ${\left( {\sqrt {\frac{x}{3}} + \frac{3}{{2{x^2}}}} \right)^{10}}$ is:

The term independent of $x$ in the expansion of $\left[\frac{x+1}{x^{2/3}-x^{1/3}+1}-\frac{x-1}{x-x^{1/2}}\right]^{10}, x \neq 1,$ is equal to ....... .

The coefficient of $x^5$ in the expansion of $(1+x^2)^5(1+x)^4$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module