Question

The AM of the series $1,2,4,8,16,.....,{{2}^{n}}$, is
A. $\dfrac{{{2}^{n}}-1}{n}$
B. $\dfrac{{{2}^{n+1}}-1}{n+1}$
C. $\dfrac{{{2}^{n+1}}-1}{n}$
D. $\dfrac{{{2}^{n}}-1}{n+1}$

Answer 1

Hint: We know that AM i.e. arithmetic mean is defined as the sum of the terms of series divided by the total number of terms. If we have ${{x}_{1}},{{x}_{2}},{{x}_{3}},......{{x}_{n}}$ terms the AM is given by;
$AM=\dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}+......+{{x}_{n}}}{n}$

Complete step-by-step solution
We have been asked to find the AM of series $1,2,4,8,16,.....,{{2}^{n}}$.
Since, we know that Am i.e. arithmetic mean is defined as sum of the terms of series divided by total number of terms. Let us suppose if we have ‘n’ terms like ${{x}_{1}},{{x}_{2}},{{x}_{3}},......{{x}_{n}}$then;
$\begin{align}
  & AM=\dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}+......+{{x}_{n}}}{n} \\
 & \Rightarrow AM=\dfrac{1+2+4+8+16+{{........2}^{n}}}{n+1} \\
\end{align}$
Here, the total number of terms is $\left( n+1 \right)$.
We can see the numerator is forming a geometric progression with first term a = 1 and common ratio r = 2.
As, we know that sum of geometric progression is given by;
${{S}_{n}}=\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1}$
 So, by applying the sum formula of G.P, we get;
$\begin{align}
  & AM=\dfrac{{{2}^{0}}+{{2}^{1}}+{{2}^{2}}+{{2}^{3}}+{{........2}^{n}}}{n+1} \\
 & =\dfrac{1\left( {{2}^{n+1}}-1 \right)}{n+1} \\
 & =\dfrac{{{2}^{n+1}}-1}{n+1} \\
\end{align}$
Therefore, the correct option is B.

Note: It is very important to observe that the total number of terms present in the series is (n+1). Sometimes by just seeing the general term i.e. ${{2}^{n}}$, we take the total number of terms as ‘n’ which is wrong as we can see. So, be careful while substituting the value of the total number of terms in the formula.