Question

Let $a=min\{x^2+2x+3,xϵR\}$ 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. $\frac{2^{n+1}-1}{{3.2}^n}$
• B. $\frac{2^{n+1}+1}{{3.2}^n}$
• C. $\frac{4^{n+1}-1}{{3.2}^n}$
• D. none of these

Answer 1

Answer: (C)

$\frac{4^{n+1}-1}{{3.2}^n}$

$x^2+2x+3={(x+1)}^2+2\ge 2$
So $a=2$
$b={lim}_{\theta \to 0}\frac{1-\cos \theta }{{\theta }^2}={lim}_{\theta \to 0}\frac{2{\sin }^2\frac{\theta }{2}}{{\theta }^2}=\frac{1}{2}$
Therefore,
${\sum }_{r=0}^n{a}^r{b}^{n-r}=b^n+a{n}^{n-1}+a^2{b}^{n-2}+...+a^n=\frac{b^n(1-{\left(\frac{a}{b}\right)}^{n+1})}{1-\frac{a}{b}}=\frac{{\left(\frac{1}{2}\right)}^n(1-4^{n+1})}{1-4}=\frac{4^{n+1}-1}{3\times 2^n}$
Hence, option 'C' is correct.