A sequence $t_{1}, t_{2}, t_{3}, \ldots$ is defined by

$$
t_{n}=\left\{\begin{array}{cl}
\frac{1}{7^{n}} & \text { when } n \text { is odd } \\
\frac{2}{7^{n}} & \text { when } n \text { is even }
\end{array}\right.
$$

for each positive integer $n$. Determine the sum of all of the terms in this sequence; that is, calculate $t_{1}+t_{2}+t_{3}+\cdots$.
We will require the following fact: If $a$ and $r$ are real numbers with $-1<r<1$, then the sum of the geometric series

$$
a+a r+a r^{2}+a r^{3}+a r^{4}+\cdots
$$

is $\frac{a}{1-r}$.

We will compute this sum by finding the sums $t_{1}+t_{3}+t_{5}+\cdots$ and $t_{2}+t_{4}+t_{6}+\cdots$ and adding the results. For the first sum, we have

$$
\frac{1}{7}+\frac{1}{7^{3}}+\frac{1}{7^{5}}+\cdots
$$

which can be rewritten as

$$
\frac{1}{7}+\frac{1}{7}\left(\frac{1}{49}\right)+\frac{1}{7}\left(\frac{1}{49}\right)^{2}+\frac{1}{7}\left(\frac{1}{49}\right)^{3}+\cdots
$$

since $7^{2}=49$.

This is a geometric series with $a=\frac{1}{7}$ and $r=\frac{1}{49}$. We also have that $-1<\frac{1}{49}<1$, so the sum of the series above is

$$
\frac{\frac{1}{7}}{1-\frac{1}{49}}=\frac{\frac{49}{7}}{49-1}=\frac{7}{48}
$$

We now compute $t_{2}+t_{4}+t_{6}+\cdots$. Similar to before, we can write the sum as

$$
\frac{2}{49}+\frac{2}{49}\left(\frac{1}{49}\right)+\frac{2}{49}\left(\frac{1}{49}\right)^{2}+\frac{2}{49}\left(\frac{1}{49}\right)^{3}+\cdots
$$

which is a geometric series with $a=\frac{2}{49}$ and $r=\frac{1}{49}$. Thus, its sum is

$$
\frac{\frac{2}{49}}{1-\frac{1}{49}}=\frac{2}{49-1}=\frac{2}{48}
$$

The sum $t_{1}+t_{2}+t_{3}+\cdots$ is therefore $\frac{7}{48}+\frac{2}{48}=\frac{9}{48}=\frac{3}{16}$.

ANSWER: $\frac{3}{16}$