Question

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

• A. $\frac{2^{n+1}-1}{3\times 2^n}$
• B. $\frac{2^{n+1}+1}{3\times 2^n}$
• C. $\frac{4^{n+1}-1}{3\times 2^n}$
• D. None of these

Answer 1

Answer: (C)

$a=\min \left\{x^2+2x+3,x\in R\right\}$
$=\min \left\{(x+1{)}^2+2,x\in R\right\}=2$ and
$b=\underset{\theta \to 0}{\lim }\frac{1-\cos \theta }{{\theta }^2}$
$=\underset{\theta \to 0}{\lim }\frac{(1-\cos \theta )(1+\cos \theta )}{{\theta }^2(1+\cos \theta )}=\frac{1}{2}$
$\therefore \sum _{r=0}^n{a}^r\cdot b^{n-r}=b^n\sum _{r=0}^n{\left(\frac{a}{b}\right)}^r={\left(\frac{1}{2}\right)}^n\sum _{r=0}^n(4{)}^r$
$=\frac{1}{2^n}\left(1+4+4^2+\dots +4^n\right)$
$=\frac{1}{2n}\cdot 1\cdot \left(\frac{4^{n+1}-1}{4-1}\right)$
$=\frac{4^{n+1}-1}{3\cdot 2^n}$