If x is positive,the first negative term in t | vedclass

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

Question

(C) The general term $T_{r+1}$ in the expansion of $(1+x)^n$ is given by $T_{r+1} = \frac{n(n-1)(n-2)\dots(n-r+1)}{r!} x^r$.For the term to be negativ

Vedclass · Accepted Answer

$8^{th}$ term

Vedclass · Answer

(C) The general term $T_{r+1}$ in the expansion of $(1+x)^n$ is given by $T_{r+1} = \frac{n(n-1)(n-2)\dots(n-r+1)}{r!} x^r$.For the term to be negative,the product $n(n-1)(n-2)\dots(n-r+1)$ must be negative since $x^r$ and $r!$ are positive for $x > 0$.Here $n = \frac{27}{5} = 5.4$.The terms become negative when the factor $(n-r+1) $5.4 - r + 1 Since $r$ must be an integer,the smallest integer $r$ is $7$.Thus,the first negative term is $T_{7+1} = T_8$,which is the $8^{th}$ term.

If $x$ is positive,the first negative term in the expansion of $(1 + x)^{27/5}$ is

Similar Questions

If $x=\frac{2 \cdot 5}{3 \cdot 6}-\frac{2 \cdot 5 \cdot 8}{3 \cdot 6 \cdot 9}\left(\frac{2}{5}\right)+\frac{2 \cdot 5 \cdot 8 \cdot 11}{3 \cdot 6 \cdot 9 \cdot 12}\left(\frac{2}{5}\right)^2-\ldots \infty$,then $7^2(12 x+55)^3=$

In the expansion of $\frac{2x+1}{(1+x)(1-2x)}$,the sum of the coefficients of the first $5$ odd powers of $x$ is

The coefficient of $x^3$ in the expansion of $(1-\frac{3}{4} x)^{\frac{1}{2}}$ is

If $(a+bx)^{-3} = \frac{1}{27} + \frac{1}{3}x + \dots$,then the ordered pair $(a, b)$ is equal to

For $n, p \in N-\{1\}$,the coefficient of $x^3$ in $\frac{(1-x)^{-1 / p}}{(1-x)^n}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module