Assertion $(A)$: If $|x| < 1$,then $\sum_{n=0}^{\infty}(-1)^n x^{n+1} = \frac{x}{x+1}$.
Reason $(R)$: If $|x| < 1$,then $(1+x)^{-1} = 1-x+x^2-x^3+\dots$.
Which one of the following is true?
Reason $(R)$: If $|x| < 1$,then $(1+x)^{-1} = 1-x+x^2-x^3+\dots$.
Which one of the following is true?
- A$A$ and $R$ are true,$R$ is a correct explanation of $A$.
- B$A$ and $R$ are true but $R$ is not a correct explanation of $A$.
- C$A$ is true,but $R$ is false.
- D$A$ is false,but $R$ is true.
Explore More
Similar Questions
Assuming $x$ to be so small that $x^2$ and higher powers of $x$ can be neglected,the coefficient of $x$ in $\frac{(1-x)^{1/3}+(1-5x)^2}{(16-x)^{1/4}}$ is equal to
If $(-c, c)$ is the set of all values of $x$ for which the expansion of $(7-5x)^{-2/3}$ is valid,then $5c + 7 =$
$1+\frac{2}{4}+\frac{2 \cdot 5}{4 \cdot 8}+\frac{2 \cdot 5 \cdot 8}{4 \cdot 8 \cdot 12}+\frac{2 \cdot 5 \cdot 8 \cdot 11}{4 \cdot 8 \cdot 12 \cdot 16}+\ldots \ldots$ is equal to :
DifficultAP EAMCET 2006
View SolutionIf the set of all values of $x$ for which the expansion of $(7-5 x)^{-\frac{2}{3}}$ is valid is equal to $(-a, a)$,then $5 a+7$ is equal to
The expression $\frac{1}{\sqrt{5 + 4x}}$ can be expanded by the binomial theorem,if
Easy
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo