How to write sum of fractions with powers

Question

Disclaimer: I'm not a mathematician so this question may sound very easy to you. Sorry if this is too easy for you. (Also english is not my native language but I'll try.)

I have the folowing formula: $$\frac{x*3^k}{(2n)^k}+\frac{3^{k−1}}{(2n)^k}+\frac{3^{k−2}}{(2n)^{k−1}}+...+\frac{3^{0}}{(2n)^{1}}$$ $$k \in \Bbb N,\;n \in \Bbb N,\;x \in \Bbb N$$
The question seems simple to me: How do I / Is there another way to write this formula, so I don't have the dots in the middle?
I you have any questions regarding this formula, please let me know, so I can clarify what I meant.

Any help is appreciated

Answer 1

You can write it as:

$\frac{x*3^k}{(2n)^k}+\frac{3^{k−1}}{(2n)^k}+\frac{3^{k−2}}{(2n)^{k−1}}+...+\frac{3^{0}}{(2n)^{1}}=$

$\frac{x*3^k}{(2n)^k}+(\frac{3^{k−1}}{(2n)^k}+\frac{3^{k−2}}{(2n)^{k−1}}+...+\frac{3^{0}}{(2n)^{1}})=$

$\frac{x*3^k}{(2n)^k}+(\frac{3^{k−1}}{(2n)^k}+\frac{3^{k−2}}{(2n)^{k−1}}+... + \frac{3^{i−1}}{(2n)^{i}}+....+\frac{3^{0}}{(2n)^{1}})=$

$\frac{x*3^k}{(2n)^k}+\sum_{i=1}^k\frac{3^{i−1}}{(2n)^i}$

In your original phrase "$\frac{x*3^k}{(2n)^k}+\frac{3^{k−1}}{(2n)^k}+\frac{3^{k−2}}{(2n)^{k−1}}+...+\frac{3^{0}}{(2n)^{1}}$" the first term does not match the rest of the others in patterns so it was isolated and written separately.

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For the heck of it, we could write the $\frac{3^{i−1}}{(2n)^i}$ terms as $\frac 13(\frac 3{2n})^i$ so the whole thing is $\frac{x*3^k}{(2n)^k}+\frac 13\sum_{i=1}^k(\frac 3{2n})^i$ if that makes anything clearer, but we don't have to (and maybe it doesn't).

As John hughes points out in his answer $\sum_{i=1}^k(\frac 3{2n})^i$ is a geometric series and you may (or may not know) $\sum_{i=1}^k(\frac 3{2n})^i= \frac{1-(\frac 3{2n})^{k+1}}{1-\frac 3{2n}}$ so the whole thing is $\frac{x*3^k}{(2n)^k}+\frac 13\frac{1-(\frac 3{2n})^{k+1}}{1-\frac 3{2n}}$.

But your question was about notation, not solving.

In notatation you can write $a_k + a_{k-1} + ..... + a_1$ as $\sum_{i=1}^k a_i$.

Answer 2

As written, we can say $$ \frac{x\cdot 3^k}{(2n)^k}+\frac{3^{k−1}}{(2n)^k}+\frac{3^{k−2}}{(2n)^{k−1}}+...+\frac{3^{0}}{(2n)^{1}} = \left(\frac{x\cdot 3^k}{(2n)^k}\right) +U $$ where \begin{align} U &= \frac{3^{k−1}}{(2n)^k}+\frac{3^{k−2}}{(2n)^{k−1}}+...+\frac{3^{0}}{(2n)^{1}}\\ &= \frac{1}{3}\frac{3^{k}}{(2n)^k}+\frac{1}{3}\frac{3^{k−1}}{(2n)^{k−1}}+...+\frac{1}{3}\frac{3^{1}}{(2n)^{1}}\\ &= \frac{1}{3}\left( \frac{3^{k}}{(2n)^k}+\frac{3^{k−1}}{(2n)^{k−1}}+...+\frac{3^{1}}{(2n)^{1}}\right)\\ &= \frac{1}{3}\left( \left(\frac{3}{(2n)}\right)^k+\ldots + \left(\frac{3}{(2n)}\right)^1\right)\\ \end{align} That middle thing is a geometric series, whose sum is $\frac{\frac{3}{2n}^{k+1}}{1 - \frac{3}{2n}} - 1$, so that the whole thing is $$ \frac{x\cdot 3^k}{(2n)^k}) + \frac{1}{3} \left( \frac{ \left(\frac{3}{2n}\right)^{k+1}}{1 - \frac{3}{2n}} - 1 \right). $$