How do you use summation notation to express the sum $32+24+18+.....+10.125$?

Hint: From the given series of geometric sequences, we find the general term of the series. We find the formula for ${{t}_{n}}$, the ${{n}^{th}}$ term of the series and use the summation notation on the general term. From the given sequence we find the common ratio which is the ratio between two consecutive terms. We also use the summation formula and put the values.

Complete step by step solution:
We have been given a series of geometric sequence which is $32+24+18+.....+10.125$
We express the geometric sequence in its general form.
We express the terms as ${{t}_{n}}$, the ${{n}^{th}}$ term of the series.
The first term be ${{t}_{1}}$ and the common ratio be $r$ where $r=\dfrac{{{t}_{2}}}{{{t}_{1}}}=\dfrac{{{t}_{3}}}{{{t}_{2}}}=\dfrac{{{t}_{4}}}{{{t}_{3}}}$.
We can express the general term ${{t}_{n}}$ based on the first term and the common ratio.
The formula being ${{t}_{n}}={{t}_{1}}{{r}^{n-1}}$.
The first term is 32. So, ${{t}_{1}}=32$. The common ratio is $r=\dfrac{24}{32}=\dfrac{18}{24}=\dfrac{3}{4}$.
We put the values of ${{t}_{1}}$ and $r$ to find the general form.
We express general term ${{t}_{n}}$ as ${{t}_{n}}={{t}_{1}}{{r}^{n-1}}=32\times {{\left( \dfrac{3}{4} \right)}^{n-1}}$.
The last term is $10.125$ which can be expressed as $10.125=32\times {{\left( \dfrac{3}{4} \right)}^{5-1}}={{t}_{5}}$.
The summation notation for the series $32+24+18+.....+10.125$ will be $\sum\limits_{n=1}^{5}{32\times {{\left( \dfrac{3}{4} \right)}^{n-1}}}$.
The value of $\left| r \right|<1$ for which the sum of the first n terms of an G.P. is ${{S}_{n}}={{t}_{1}}\dfrac{1-{{r}^{n}}}{1-r}$.
Now we place $n=5$ to get the sum where ${{S}_{5}}=32\times \dfrac{1-{{\left( \dfrac{3}{4} \right)}^{5}}}{1-\dfrac{3}{4}}$. Simplified form will be
${{S}_{5}}=32\times \dfrac{1-{{\left( \dfrac{3}{4} \right)}^{5}}}{1-\dfrac{3}{4}}=128\times \text{0}\text{.7626953125}=97.625$.

Note: The sequence is an increasing sequence where the common ratio is a positive number. The common difference will never be calculated according to the difference of greater number from the lesser number. The ratio formula should always be according $r=\dfrac{{{t}_{2}}}{{{t}_{1}}}=\dfrac{{{t}_{3}}}{{{t}_{2}}}=\dfrac{{{t}_{4}}}{{{t}_{3}}}$.