Question

Let $a=min\{x^2+2x+3,x\in R\}$ 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)

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

Let $a=min\{x^2+2x+3:x\in R\}$ and $b=\underset{\theta \to 0}{lim}\frac{1-\cos \theta }{{\theta }^2}$

Now, $x^2+2x+3=x^2+2x+1+2=(x+1{)}^2+2$

$\therefore min\{x^2+2x+3:x\in R\}=min\{{(x+1)}^2+2:x\in R\}=2$

Also, $b=\underset{\theta \to 0}{lim}\frac{1-\cos \theta }{{\theta }^2}$

$=\underset{\theta \to 0}{lim}\frac{\sin \theta }{2\theta }=\frac{1}{2}$ ...... $\left\{\underset{\theta \to 0}{lim}\frac{\sin \theta }{\theta }=1\right\}$

$\sum _{r=0}^n{a}^r\cdot b^{n-r}$

$=\sum _{r=0}^n{2}^r{\left(\frac{1}{2}\right)}^{n-r}=2^{-n}{\sum }_{r=0}^n{2}^{2r}=2^{-n}\{1+2^2+2^4{+.....2}^{2n}\}=2^{-n}.\frac{1(4^{n+1}-1)}{4-1}=\frac{4^{n+1}-1}{{3.2}^n}$