Binomial-Theorem-And-Its-Simple-Applications Question 237
Question: Which of the following result is valid?
Options:
A) $ {{(1+x)}^{n}}>(1+nx), $ For all natural numbers n
B) $ {{(1+x)}^{n}}\ge (1+nx), $ For all natural numbers n, Where $ x>-1 $
C) $ {{(1+x)}^{n}}\le (1+nx), $ For all natural numbers n
D) $ {{(1+x)}^{n}}<(1+nx), $ For all natural numbers n
Correct Answer: BShow Answer
Answer:
Solution:
For $ n=1,{{(1+x)}^{1}}=1+x $
$ =1+1x\ge 1+1.x{{(1+x)}^{1}}\ge 1+1.x $
For $ n=k, $ let $ P(k):{{(1+x)}^{k}}\ge (1+kx) $ is true.
For $ n=k+1,P(k+1):{{(1+x)}^{k+1}}\ge {1+(k+1)x} $
is also true.
We will show $ P(k+1) $ is true.
Consider $ {{(1+x)}^{k+1}}={{(1+x)}^{k}}(1+x)\ge (1+kx)(1+x) $ $ [ifx>-1] $
$ =1+x+kx+kx^{2}\ge 1+x+kx $ $ [\because ,k>0andx>-1] $
$ =1+(k+1)x $
Thus, $ {{(1+x)}^{k+1}}\ge 1+(k+1)x,ifx>-1 $
BETA
