Question

The geometric mean of $1,2,{{2}^{2}},..........,{{2}^{n}}$ is:
(1) ${{2}^{\dfrac{n}{2}}}$
(2) ${{n}^{\dfrac{\left( n+1 \right)}{2}}}$
(3) ${{2}^{\dfrac{n\left( n+1 \right)}{2}}}$
(4) ${{2}^{\dfrac{\left( n-1 \right)}{2}}}$

Answer 1

Hint: Here in this question we have been asked to find the geometric mean of $1,2,{{2}^{2}},..........,{{2}^{n}}$. From the basic concepts, we know that the geometric mean of ${{x}_{1}},{{x}_{2}},..........,{{x}_{n}}$ is given as $\sqrt[n]{{{x}_{1}}{{x}_{2}}..........{{x}_{n}}}$. We will use the formula for finding the sum of the first $n$ natural numbers are given as $\dfrac{n\left( n+1 \right)}{2}$ .

Complete step by step solution:
Now considering from the question we have been asked to find the geometric mean of $1,2,{{2}^{2}},..........,{{2}^{n}}$.
From the basic concepts, we know that the geometric mean of ${{x}_{1}},{{x}_{2}},..........,{{x}_{n}}$ is given as $\sqrt[n]{{{x}_{1}}{{x}_{2}}..........{{x}_{n}}}$.
Hence we can say that the geometric mean of $1,2,{{2}^{2}},..........,{{2}^{n}}$ will be given as $\sqrt[n+1]{{{1.2.2}^{2}}{{......2}^{n}}}$ .
We will use the formula for finding the sum of the first $n$ natural numbers given as $\dfrac{n\left( n+1 \right)}{2}$ which we have learnt from the basics.
Let us simplify the expression we have using the above formula.
By doing that we will have
 $\begin{align}
  & \Rightarrow \sqrt[n+1]{{{2}^{0+1+2+.....+n}}} \\
 & \Rightarrow \sqrt[n+1]{{{2}^{\left( \dfrac{n\left( n+1 \right)}{2} \right)}}} \\
\end{align}$ .
Now by further simplifying this we will have $ {{2}^{\dfrac{n}{2}}}$ .
Therefore we can conclude that the geometric mean of $1,2,{{2}^{2}},..........,{{2}^{n}}$will be given as ${{2}^{\dfrac{n}{2}}}$ .
Hence we will mark the option “1” as correct.

Note: In the process of answering questions of this type, we should be sure with the concepts that we are going to apply in between the steps. This is a very simple question in which very few mistakes are possible and this can be answered in a short span of time.