Using the Binomial Theorem,determine which number is larger: $(1.1)^{10000}$ or $1000$.
(A) Using the Binomial Theorem,we expand $(1.1)^{10000}$ as follows:
$(1.1)^{10000} = (1 + 0.1)^{10000}$
$= {^{10000}}C_0 (1)^{10000} + {^{10000}}C_1 (1)^{9999} (0.1) + \text{other positive terms}$
$= 1 + 10000 \times 0.1 + \text{other positive terms}$
$= 1 + 1000 + \text{other positive terms}$
$= 1001 + \text{other positive terms}$
Since all terms in the expansion are positive,it is clear that $1001 + \text{other positive terms} > 1000$.
Therefore,$(1.1)^{10000} > 1000$.
$(1.1)^{10000} = (1 + 0.1)^{10000}$
$= {^{10000}}C_0 (1)^{10000} + {^{10000}}C_1 (1)^{9999} (0.1) + \text{other positive terms}$
$= 1 + 10000 \times 0.1 + \text{other positive terms}$
$= 1 + 1000 + \text{other positive terms}$
$= 1001 + \text{other positive terms}$
Since all terms in the expansion are positive,it is clear that $1001 + \text{other positive terms} > 1000$.
Therefore,$(1.1)^{10000} > 1000$.
Explore More
Similar Questions
$(1 + x)^n - nx - 1$ is divisible by (where $n \in N$):
Easy
View SolutionIf $|x|$ is so small that all terms containing $x^2$ and higher powers of $x$ can be neglected,then the approximate value of $\frac{(3-5x)^{1/2}}{(5-3x)^2}$,when $x=\frac{1}{\sqrt{363}}$,is
If the roots of $x^5-ax^4+bx^3-cx^2+dx-1=0$ are all positive such that their arithmetic mean and geometric mean are equal,then $a+b+c+d=$
Using the binomial theorem, the value of $(0.999)^3$ correct to $3$ decimal places is:
MediumWBJEE 2009
View SolutionIf the sum of the coefficients in the expansion of $(\alpha x^2 - 2x + 1)^{35}$ is equal to the sum of the coefficients in the expansion of $(x - \alpha y)^{35}$,then $\alpha = $
Difficult
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