Differentiation Question 167
Question: If $ a=\min {x^{2}+4x+5,x\in R} $ and $ b=\underset{\theta \to 0}{\mathop{\lim }}\frac{1-\cos 2\theta }{{{\theta }^{2}}}, $ then the value of $ \sum\limits _{r=0}^{n}{a^{r}.{b^{n-r}}} $ is
Options:
A) $ \frac{{2^{n+1}}-1}{{{4.2}^{n}}} $
B) $ {2^{n+1}}-1 $
C) $ \frac{{2^{n+1}}-1}{{{3.2}^{n}}} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ x^{2}+4x+5={{(x+2)}^{2}}+1\ge 1. $ so, $ a=1 $
$ b=\underset{\theta \to 0}{\mathop{\lim }}\frac{2{{\sin }^{2}}\theta }{{{\theta }^{2}}}=2 $
$ \sum\limits _{r=0}^{n}{a^{r}.{b^{n-r}}=b^{n}+a{b^{n-1}}}+a^{2}{b^{n-2}}+…+a^{n} $
$ =\frac{b^{n}[ 1-{{( \frac{a}{b} )}^{n+1}} ]}{1-\frac{a}{b}}=\frac{2^{n}[ 1-{{( \frac{1}{2} )}^{n+1}} ]}{1-\frac{1}{2}} $
$ =\frac{{2^{n+1}}({2^{n+1}}-1)}{{2^{n+1}}}=({2^{n+1}}-1). $
BETA
