Question

Let $a=min\{x^2+2x+3:x\in R\}$ and $b=\underset{\theta \to 0}{lim}\frac{1-\cos \theta }{{\theta }^2}.$ Then $\sum _{r=0}^n{a}^r{b}^{n-r}$ is

• A. $\frac{2^{n+1}-1}{3\cdot 2^n}$
• B. $\frac{2^{n+1}+1}{3\cdot 2^n}$
• C. $\frac{4^{n+1}-1}{3\cdot 2^n}$
• D. $\frac{1}{2}(2^n-1)$

Answer 1

The correct option is C $\frac{4^{n+1}-1}{3\cdot 2^n}$

$(x^2+2x+3)=(x+1{)}^2+2\Rightarrow min(x^2+2x+3)=2\Rightarrow a=2$

$b=\underset{\theta \to 0}{lim}\frac{1-\cos \theta }{{\theta }^2}$
Using L'Hospital
$=\underset{\theta \to 0}{lim}\frac{\sin \theta }{2\theta }$
$=\frac{1}{2}$

Now,
$\sum _{r=0}^n{a}^r{b}^{n-r}$
$=b^n\sum _{r=0}^n{\left(\frac{a}{b}\right)}^r$
$=\frac{1}{2^n}\cdot \sum _{r=0}^n{4}^r$
$=\frac{1}{2^n}(1+4+4^2+...+4^n)$
$=\frac{1}{2^n}\left(\frac{4^{n+1}-1}{4-1}\right)$
$=\frac{4^{n+1}-1}{3\cdot 2^n}$