College Algebra, Chapter 9, 9.2, Section 9.2, Problem 56

Determine the product of the numbers $10^{\frac{1}{10}}, 10^{\frac{2}{10}}, 10^{\frac{3}{10}}, 10^{\frac{4}{10}},....., 10^{\frac{19}{10}}$

By Laws of Exponent, we have

$10^{\frac{1}{10} + \frac{2}{10} + \frac{3}{10}, \frac{4}{10} + ..... + \frac{19}{10}}$

To solve for the sum, we use both formulas of partial sums of the arithmetic sequence; solve for $n$, where $\displaystyle d = \frac{2}{10} -\frac{1}{10} = \frac{1}{10}$


$\begin{equation} \begin{aligned} \frac{n}{2} \left[ 2a + (n - 1) d \right] =& n \left( \frac{a + a_n}{2} \right) && \\ \\ 2a + (n - 1)d =& a + a_n && \text{Multiply both sides by } \frac{2}{n} \\ \\ (n - 1)d =& a_n - a && \text{Combine like terms} \\ \\ n - 1 =& \frac{a_n - a}{d} && \text{Divide by } d \\ \\ n =& \frac{a_n - a}{d} + 1 && \text{Add } 1 \\ \\ n =& \frac{\displaystyle \frac{19}{10} - \frac{1}{10}}{ \displaystyle \frac{1}{10}} + 1 && \\ \\ n =& \frac{\displaystyle \frac{18}{\cancel{10}}}{\displaystyle \frac{1}{\cancel{10}}} + 1 && \\ \\ n =& 19 && \end{aligned} \end{equation}$


Now we solve the partial sum,


$\begin{equation} \begin{aligned} S_{19} =& 19 \left( \frac{\displaystyle \frac{1}{10} + \frac{19}{10}}{2} \right) \\ \\ S_{19} =& 19(1) \\ \\ S_{19} =& 19 \end{aligned} \end{equation}$


So the product is

$10^{\frac{1}{10} + \frac{2}{10} + \frac{3}{10} + \frac{4}{10} + ..... + \frac{19}{10}} = 10^{19}$