Question

Find the A.M. of the series \[1,2,4,8,16,...,{{2}^{n}}\].
A. $\dfrac{{{2}^{n+1}}-1}{n+1}$
B. $\dfrac{{{2}^{n+2}}-1}{n}$
C. $\dfrac{{{2}^{n}}-1}{n+1}$
D. $\dfrac{{{2}^{n}}-1}{n}$

Answer 1

Hint: We will first observe this series and see if it is forming any kind of progression like AP, GP or HP. We will observe that it is forming a GP. Then we will use the formula $AM=\dfrac{\text{sum of series}}{\text{total number of terms in the series}}$ to find the AM. We will find the sum of the series by using the formula for sum of a GP given as $\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1}$ where a is the first term of the series, r is the common ratio and n is the total number of terms in the series. We will also be able to observe that the total number of terms in the given series is n+1. Then we will put all these values in the formula for AM and hence we will get our answer.

Complete step-by-step answer:
Here, we have been given the series $1,2,4,8,16,...,{{2}^{n}}$ and we have to find its A.M.
If we observe this series, we can see that:
$\begin{align}
  & {{2}^{0}}=1 \\
 & {{2}^{1}}=2 \\
 & {{2}^{2}}=4 \\
 & {{2}^{3}}=8 \\
 & {{2}^{4}}=16 \\
\end{align}$
Thus, this series can be written as:
${{2}^{0}},{{2}^{1}},{{2}^{2}},{{2}^{3}},{{2}^{4}},...,{{2}^{n}}$
Now we can see that this series is a geometric progression. Now we will find its A.M.
We now know that the AM of any series is given as:
$AM=\dfrac{\text{sum of series}}{\text{total number of terms in the series}}$
Now in this series, we can see that the first term in this series ${{2}^{0}}$ and the last term is ${{2}^{n}}$. Hence, the total number of terms in this series is n+1.
Now, we will calculate the sum of the series.
This term is a GP and we know the sum of a GP is given as: $\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1}$ where a is the first term of the series, r is the common ratio and n is the total number of terms in the series.
Now, we have been given the series $1,2,4,8,16,...,{{2}^{n}}$ which can also be written as ${{2}^{0}},{{2}^{1}},{{2}^{2}},{{2}^{3}},{{2}^{4}},...,{{2}^{n}}$.
Now, we can see that in this series,
a=1
r=2
n=n+1
Now, putting these values in the formula for the sum, we get:
 $\begin{align}
  & \dfrac{a\left( {{r}^{n}}-1 \right)}{r-1} \\
 & \Rightarrow \dfrac{1\left( {{2}^{n+1}}-1 \right)}{2-1} \\
 & \Rightarrow {{2}^{n+1}}-1 \\
\end{align}$
Thus, the sum of the series is ${{2}^{n+1}}-1$.
Now, we have already established above that the total number of terms of the series is n+1.
Thus, we have our AM as:
$\begin{align}
  & AM=\dfrac{\text{sum of series}}{\text{total number of terms in the series}} \\
 & \Rightarrow AM=\dfrac{{{2}^{n+1}}-1}{n+1} \\
\end{align}$
Thus, the AM of the series is $\dfrac{{{2}^{n+1}}-1}{n+1}$.

So, the correct answer is “Option A”.

Note: Here, we have used the sum of GP. It will come in handy if we know something about GPs and APs.
GP is a series in which the ratio between every successive term is equal.
A general GP is represented as:
$a,ar,a{{r}^{2}},a{{r}^{3}},....$
General term: ${{T}_{n}}=a{{r}^{n-1}}$
Sum of n terms: $\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1}$
AP is a series in which the difference between every successive term is equal.
A general AP is represented as:
$a,a+d,a+2d,a+3d,...$
General term: ${{T}_{n}}=a+\left( n-1 \right)d$
Sum of n terms: $\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)$