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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

Last updated date: 19th Jul 2024
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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,
  & 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) \\
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,
  & 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) \\
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}}}} \\
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,
  & \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) \\
 & \\
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,
  & \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} \\

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.