How can I write the following series in summation notation?

Question

I am writing up a report for my linear algebra class where I am using series to describe certain transformations in a particular vector space:

One example is:

$$t_2 + t_5 + t_8 + ... + t_{n-1} = \sum_{i=0}^{k-1} t_{2+3i}$$

I would like to write:

$$t_1+t_2+t_4+t_5+t_7+t_8+...+t_n$$ in summation notation as well, but I am unsure how to approach it. The next subscript alternates between $+1$ or $+2$ to the previous term's subscript and is guaranteed to end in $n$. In this case, it is guaranteed that $n \equiv 1\pmod{3}$.

Answer 1

One fairly simple way to approach this is to note you're requesting the summation of all of the terms from $t_1$ to $t_n$, apart from those with indices which are integral multiples of $3$, but with a guarantee to always include $t_n$. This becomes

$$\begin{equation}\begin{aligned} S & = t_1 + t_2 + t_4 + t_5 + t_7 + t_8 + \ldots + t_n \\ & = (t_1 + t_2 + \ldots + t_n) - \left(t_3 + t_6 + \ldots + t_{3\left\lfloor\frac{n-1}{3}\right\rfloor}\right) \\ & = \sum_{i=1}^{n}t_i - \sum_{i=1}^{\left\lfloor\frac{n-1}{3}\right\rfloor}t_{3i} \end{aligned}\end{equation}$$

For your specific case that $n \equiv 1\pmod{3}$, then $\left\lfloor\frac{n-1}{3}\right\rfloor=\frac{n-1}{3}$, so the above equation simplifies to

$$S = \sum_{i=1}^{n}t_i - \sum_{i=1}^{\frac{n-1}{3}}t_{3i}$$

We could instead have just one summation by including that the indices $i$ are not integral multiples of $3$, i.e., $3 \nmid i$, in the summation conditions to then get

$$S = \sum_{i=1,\,3\,\nmid\,i}^{n}t_i$$

Answer 2

Knowing that $n \equiv 1 \pmod3,$ another way to write the sum is

$$ t_1 + \sum_{i=1}^{(n-1)/3} (t_{3i - 1} + t_{3i + 1}). $$

Equivalently,

$$ \left(\sum_{i=0}^{(n-4)/3} (t_{3i+1} + t_{3i+2})\right) + t_n. $$

The reason for the extra term outside the sum is that while the alternating $+1$ and $+2$ differences cause your terms $t_k$ to naturally occur in pairs, the stipulation $n \equiv 1 \pmod3$ means you have an odd number of terms. If you had required that $n \equiv 2 \pmod3$ all the terms would have been included inside the summation notation.