Question

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

Answer 1

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

In the question, it is given a=\min \left\\{ {{x}^{2}}+2x+3,x\in R \right\\}
$\Rightarrow a=$ minimum value of ${{x}^{2}}+2x+3$
For a quadratic polynomial $a{{x}^{2}}+bx+c$, the minimum value is given by the formula,
$\dfrac{-\left( {{b}^{2}}-4ac \right)}{4a}.............\left( 1 \right)$
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,
$\begin{aligned} & a=\dfrac{-\left( {{\left( 2 \right)}^{2}}-4\left( 1 \right)\left( 3 \right) \right)}{4\left( 1 \right)} \\\ & \Rightarrow a=\dfrac{-\left( 4-12 \right)}{4} \\\ & \Rightarrow a=\dfrac{-\left( -8 \right)}{4} \\\ & \Rightarrow a=\dfrac{8}{4} \\\ & \Rightarrow a=2...........\left( 2 \right) \\\ \end{aligned}$
Also, it is given in the question $b=\underset{\theta \to 0}{\mathop{\lim }}\,\dfrac{1-\cos \theta }{{{\theta }^{2}}}$. If we substitute $\theta =0$ in the limit function, we can notice that this limit is of the form $\dfrac{0}{0}$. Since this limit is of the form $\dfrac{0}{0}$, 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,
$\begin{aligned} & b=\underset{\theta \to 0}{\mathop{\lim }}\,\dfrac{\dfrac{d\left( 1-\cos \theta \right)}{d\theta }}{\dfrac{d{{\theta }^{2}}}{d\theta }} \\\ & \Rightarrow b=\underset{\theta \to 0}{\mathop{\lim }}\,\dfrac{\sin \theta }{2\theta } \\\ & \Rightarrow b=\dfrac{1}{2}\underset{\theta \to 0}{\mathop{\lim }}\,\dfrac{\sin \theta }{\theta }...........\left( 3 \right) \\\ \end{aligned}$
There is a formula $\underset{\theta \to 0}{\mathop{\lim }}\,\dfrac{\sin \theta }{\theta }=1$. Substituting $\underset{\theta \to 0}{\mathop{\lim }}\,\dfrac{\sin \theta }{\theta }=1$ in equation $\left( 3 \right)$, we get,
$b=\dfrac{1}{2}...........\left( 4 \right)$
In the question, it is asked to find the value of $\sum\limits_{r=0}^{n}{{{a}^{r}}.{{b}^{n-r}}}$. Substituting equation $\left( 2 \right)$ and equation $\left( 4 \right)$ in $\sum\limits_{r=0}^{n}{{{a}^{r}}.{{b}^{n-r}}}$, we get,

& \sum\limits_{r=0}^{n}{{{a}^{r}}.{{b}^{n-r}}}=\sum\limits_{r=0}^{n}{{{2}^{r}}.\dfrac{1}{{{2}^{n-r}}}} \\\ & \Rightarrow \sum\limits_{r=0}^{n}{{{a}^{r}}.{{b}^{n-r}}}=\sum\limits_{r=0}^{n}{{{2}^{r-\left( n-r \right)}}} \\\ & \Rightarrow \sum\limits_{r=0}^{n}{{{a}^{r}}.{{b}^{n-r}}}=\sum\limits_{r=0}^{n}{{{2}^{r-n+r}}} \\\ & \Rightarrow \sum\limits_{r=0}^{n}{{{a}^{r}}.{{b}^{n-r}}}=\sum\limits_{r=0}^{n}{{{2}^{2r-n}}} \\\ & \Rightarrow \sum\limits_{r=0}^{n}{{{a}^{r}}.{{b}^{n-r}}}=\sum\limits_{r=0}^{n}{\dfrac{{{2}^{2r}}}{{{2}^{n}}}} \\\ & \Rightarrow \sum\limits_{r=0}^{n}{{{a}^{r}}.{{b}^{n-r}}}=\sum\limits_{r=0}^{n}{\dfrac{{{4}^{r}}}{{{2}^{n}}}} \\\ \end{aligned}$$ Since the limits of this summation is with respect to $r$, we can take $\dfrac{1}{{{2}^{n}}}$ out of the summation. $$\Rightarrow \sum\limits_{r=0}^{n}{{{a}^{r}}.{{b}^{n-r}}}=\dfrac{1}{{{2}^{n}}}\sum\limits_{r=0}^{n}{{{4}^{r}}}$$ Evaluating the summation, we get, $$\begin{aligned} & \Rightarrow \sum\limits_{r=0}^{n}{{{a}^{r}}.{{b}^{n-r}}}=\dfrac{1}{{{2}^{n}}}\left( {{4}^{0}}+{{4}^{1}}+{{4}^{2}}+{{4}^{3}}+............+{{4}^{n}} \right) \\\ & \Rightarrow \sum\limits_{r=0}^{n}{{{a}^{r}}.{{b}^{n-r}}}=\dfrac{1}{{{2}^{n}}}\left( 1+{{4}^{1}}+{{4}^{2}}+{{4}^{3}}+............+{{4}^{n}} \right).........\left( 5 \right) \\\ & \\\ \end{aligned}$$ The above series is a geometric progression of which we have to calculate the sum. The sum of the G.P. $a,ar,a{{r}^{2}},a{{r}^{3}},............,a{{r}^{x}}$ is given by the formula, $S=\dfrac{a\left( {{r}^{x}}-1 \right)}{r-1}........\left( 6 \right)$ From equation $\left( 5 \right)$, substituting $a=1,r=4,x=n+1$ in equation $\left( 6 \right)$, we get, $$\begin{aligned} & \sum\limits_{r=0}^{n}{{{a}^{r}}.{{b}^{n-r}}}=\dfrac{1}{{{2}^{n}}}\left( \dfrac{1\left( {{4}^{n+1}}-1 \right)}{4-1} \right) \\\ & \Rightarrow \sum\limits_{r=0}^{n}{{{a}^{r}}.{{b}^{n-r}}}=\dfrac{1}{{{2}^{n}}}\left( \dfrac{{{4}^{n+1}}-1}{3} \right) \\\ & \Rightarrow \sum\limits_{r=0}^{n}{{{a}^{r}}.{{b}^{n-r}}}=\dfrac{{{4}^{n+1}}-1}{{{2}^{n}}.3} \\\ \end{aligned}$$ 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.