Question

If $a=min\{x^2+4x+5,x\in R\}$ and $b=\underset{\theta \to 0}{lim}\frac{1-\cos 2\theta }{{\theta }^2}$ then the value of$\sum _{r=0}^n{a}^r{b}^{n-r}=$

• A. $\frac{2^{n+1}-1}{{4.2}^n}$
• B. $2^{n+1}-1$
• C. $\frac{2^{n+1}-1}{{3.2}^n}$
• D. $2^n-1$

Answer 1

Answer: (B)

$2^{n+1}-1$

$a=min\{x^2+4x+5,xϵR\}$

$x^2+4x+5=x^2+4x+4+1$

$={(x+2)}^2+1$

${(x+2)}^2\ge 0$

$\Rightarrow {(x+2)}^2+1\ge 1$

$\therefore$ min value of $x^2+4x+5$ is $1$

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

${\sum }_{r=0}^n{a}^r{b}^{n-r}=(2^n+2^{n-1}+2^{n-2}+...+1)$

Sum of GP$=\frac{a(1-r^n)}{(1-r)}$n$\to$no. of terms $=n+1$, r$\to$common ratio $=\frac{1}{2}$

$=\frac{2^n(1-{\left(\frac{1}{2}\right)}^{n+1})}{1-\frac{1}{2}}$

$=2^{n+1}(1-\frac{1}{2^{n+1}})$

$=2^{n+1}-1$