Question

Let $a=min\{x^2+2x+3,x\in R\}$ and $b=\underset{\theta \to 0}{lim}{\displaystyle \frac{1-\cos \theta }{{\theta }^2}}$. The value of $\sum _{r=0}^n{a}^r.b^{n-r}$ is
(a) ${\displaystyle \frac{2^{n+1}-1}{{3.2}^n}}$
(b) ${\displaystyle \frac{2^{n+1}+1}{{3.2}^n}}$
(c) ${\displaystyle \frac{4^{n+1}-1}{{3.2}^n}}$
(d) none of these

Answer 1

Hint: ${}^{\prime }{a}^{\prime }$ can be found out by using the formula for minimum value of a quadratic polynomial. We can use the L' Hopital rule to find ${}^{\prime }{b}^{\prime }$ . The required answer in the form of summation is a Geometric progression.

In the question, it is given $a=min\{x^2+2x+3,x\in R\}$
$\Rightarrow a=$ minimum value of $x^2+2x+3$
For a quadratic polynomial $a{x}^2+bx+c$, the minimum value is given by the formula,
${\displaystyle \frac{-(b^2-4ac)}{4a}}.............\left(1\right)$
Since the polynomial given in the question is $x^2+2x+3$, substituting $a=1,b=2,c=3$ in equation $\left(1\right)$, we get,
$\begin{array}{rl}& a={\displaystyle \frac{-({\left(2\right)}^2-4\left(1\right)\left(3\right))}{4\left(1\right)}}\\ & \Rightarrow a={\displaystyle \frac{-(4-12)}{4}}\\ & \Rightarrow a={\displaystyle \frac{-(-8)}{4}}\\ & \Rightarrow a={\displaystyle \frac{8}{4}}\\ & \Rightarrow a=2...........\left(2\right)\end{array}$
Also, it is given in the question $b=\underset{\theta \to 0}{lim}{\displaystyle \frac{1-\cos \theta }{{\theta }^2}}$. If we substitute $\theta =0$ in the limit function, we can notice that this limit is of the form $${\displaystyle \frac{0}{0}}$$. Since this limit is of the form $${\displaystyle \frac{0}{0}}$$, we can use L’ Hopital rule to solve this limit. In L’ Hopital rule, we individually differentiate the numerator and the denominator with respect to the limit variable i.e. $\theta$ in this case and then apply the limit again.
Applying L’ Hopital rule on $b$, we get,
$\begin{array}{rl}& b=\underset{\theta \to 0}{lim}{\displaystyle \frac{{\displaystyle \frac{d(1-\cos \theta )}{d\theta }}}{{\displaystyle \frac{d{\theta }^2}{d\theta }}}}\\ & \Rightarrow b=\underset{\theta \to 0}{lim}{\displaystyle \frac{\sin \theta }{2\theta }}\\ & \Rightarrow b={\displaystyle \frac{1}{2}}\underset{\theta \to 0}{lim}{\displaystyle \frac{\sin \theta }{\theta }}...........\left(3\right)\end{array}$
There is a formula $\underset{\theta \to 0}{lim}{\displaystyle \frac{\sin \theta }{\theta }}=1$. Substituting $\underset{\theta \to 0}{lim}{\displaystyle \frac{\sin \theta }{\theta }}=1$ in equation $\left(3\right)$, we get,
$b={\displaystyle \frac{1}{2}}...........\left(4\right)$
In the question, it is asked to find the value of $\sum _{r=0}^n{a}^r.b^{n-r}$. Substituting equation $\left(2\right)$ and equation $\left(4\right)$ in $\sum _{r=0}^n{a}^r.b^{n-r}$, we get,
$$\begin{array}{rl}& \sum _{r=0}^n{a}^r.b^{n-r}=\sum _{r=0}^n{2}^r.{\displaystyle \frac{1}{2^{n-r}}}\\ & \Rightarrow \sum _{r=0}^n{a}^r.b^{n-r}=\sum _{r=0}^n{2}^{r-(n-r)}\\ & \Rightarrow \sum _{r=0}^n{a}^r.b^{n-r}=\sum _{r=0}^n{2}^{r-n+r}\\ & \Rightarrow \sum _{r=0}^n{a}^r.b^{n-r}=\sum _{r=0}^n{2}^{2r-n}\\ & \Rightarrow \sum _{r=0}^n{a}^r.b^{n-r}=\sum _{r=0}^n{\displaystyle \frac{2^{2r}}{2^n}}\\ & \Rightarrow \sum _{r=0}^n{a}^r.b^{n-r}=\sum _{r=0}^n{\displaystyle \frac{4^r}{2^n}}\end{array}$$
Since the limits of this summation is with respect to $r$, we can take ${\displaystyle \frac{1}{2^n}}$ out of the summation.
$$\Rightarrow \sum _{r=0}^n{a}^r.b^{n-r}={\displaystyle \frac{1}{2^n}}\sum _{r=0}^n{4}^r$$
Evaluating the summation, we get,
$$\begin{array}{rl}& \Rightarrow \sum _{r=0}^n{a}^r.b^{n-r}={\displaystyle \frac{1}{2^n}}(4^0+4^1+4^2+4^3+............+4^n)\\ & \Rightarrow \sum _{r=0}^n{a}^r.b^{n-r}={\displaystyle \frac{1}{2^n}}(1+4^1+4^2+4^3+............+4^n).........\left(5\right)\\ & \end{array}$$
The above series is a geometric progression of which we have to calculate the sum.
The sum of the G.P. $a,ar,a{r}^2,a{r}^3,............,a{r}^x$ is given by the formula,
$S={\displaystyle \frac{a(r^x-1)}{r-1}}........\left(6\right)$
From equation $\left(5\right)$, substituting $a=1,r=4,x=n+1$ in equation $\left(6\right)$, we get,
$$\begin{array}{rl}& \sum _{r=0}^n{a}^r.b^{n-r}={\displaystyle \frac{1}{2^n}}\left({\displaystyle \frac{1(4^{n+1}-1)}{4-1}}\right)\\ & \Rightarrow \sum _{r=0}^n{a}^r.b^{n-r}={\displaystyle \frac{1}{2^n}}\left({\displaystyle \frac{4^{n+1}-1}{3}}\right)\\ & \Rightarrow \sum _{r=0}^n{a}^r.b^{n-r}={\displaystyle \frac{4^{n+1}-1}{2^n.3}}\end{array}$$

So the answer is option (c)

Note: There is a possibility of error while finding the value of $b$, since it involves derivative of $\cos x$ which is equal to $-\sin x$. But sometimes, we may get confused while applying the negative sign and may write the derivative of $\cos x$ as $\sin x$ which will lead to an incorrect answer.