What is the closed-form expression for $T_n = \left(\sum_{i=1}^{n-1}7 T_i\right) + 1$ where $T_1 = 1 ?$
After calculating a few values, you can guess the solution and then easily prove it by induction. Another way to find it is to use "creative cancellation":
$$1 + 7 (T_1 + \cdots + T_{n-1}) = T_n = (T_n + \cdots + T_1) - (T_{n-1} + \cdots + T_1).$$
Hence if you put $S_n = T_1 + \cdots + T_{n-1}$, you deduce
$$
S_n = 8S_{n-1} + 1.
$$
In order to make this homogeneous, take $R_n = S_n + c$:
$$
R_n = S_n + c = 8S_{n-1} + 1 + c = 8(R_{n-1} - c) + 1 + c = 8R_{n-1} + 1 - 7c.
$$
Choosing $c = 1/7$, this gives $R_n = 8R_{n-1}$, and so $R_n = 8^{n-1} R_1$. Now $R_1 = S_1 + 1/7 = T_1 + 1/7 = 8/7$, and so $R_n = 8^n/7$.

It follows that $S_n = (8^n-1)/7$ and $T_n = S_n - S_{n-1} = (8^n - 8^{n-1})/7 = 8^{n-1}$.



Let me conclude by giving a combinatorial interpretation for the recurrence. I will show that $T_n$ is the number of base-8 numbers of length $n-1$. Indeed, any such number is either 0, or starts with a non-zero digit followed by a number $i &lt; n-1$ digits long.