The approximate value of $${\left( {1.0002} \right)^{3000}}$$ is
A.
1.6
B.
1.4
C.
1.8
D.
1.2
Answer :
1.6
Solution :
$$\eqalign{
& {\left( {1.0002} \right)^{3000}} = {\left( {1 + 0.0002} \right)^{3000}} \cr
& = 1 + \left( {3000} \right)\left( {0.0002} \right) + \frac{{\left( {3000} \right)\left( {2999} \right)}}{{1.2}}{\left( {0.0002} \right)^2} + \frac{{\left( {3000} \right)\left( {2999} \right)\left( {2998} \right)}}{{1.2.3}}{\left( {0.0002} \right)^3} + ..... \cr} $$
We want to get answer correct to only one decimal places and as such we have left further expansion.
$$ = 1 + \left( {3000} \right)\left( {0.0002} \right) = 1.6$$
Releted MCQ Question on Algebra >> Binomial Theorem
Releted Question 1
Given positive integers $$r > 1, n > 2$$ and that the co - efficient of $${\left( {3r} \right)^{th}}\,{\text{and }}{\left( {r + 2} \right)^{th}}$$ terms in the binomial expansion of $${\left( {1 + x} \right)^{2n}}$$ are equal. Then
If in the expansion of $${\left( {1 + x} \right)^m}{\left( {1 - x} \right)^n},$$ the co-efficients of $$x$$ and $${x^2}$$ are $$3$$ and $$- 6\,$$ respectively, then $$m$$ is