Binomial Theorem And Its Simple Applications Question 104
Question: The approximate value of $ {{(1.0002)}^{3000}} $ is
Options:
A) 1.6
B) 1.4
C) 1.8
D) 1.2
Show Answer
Answer:
Correct Answer: A
Solution:
- [a] $ {{(1.0002)}^{3000}}={{(1+0.0002)}^{3000}} $
$ =1+(3000)(0.0002)+\frac{(3000)(2999)}{1.2}{{(0.0002)}^{2}}+ $
$ \frac{(3000)(2999)(2998)}{1.2.3}{{(0.0002)}^{3}}+…. $ We want to get answer correct to only one decimal place and as such we have left further expansion. $ =1+(3000)(0.0002)=1.6 $
BETA
