I got a sequence where the relation between elements of the sequence is given by: \begin{align} y_1 &= b \\ y_{i+1} &= 2 y_i + b \quad (i \in \mathbb{N}) \end{align} where $b$ is called base, $y$ is the sequence value, and $i$ is the $i$-th element of the sequence.
For $b=1$ we could write $y_i$ as $$ y_i = 2^i - 1 $$
The threshold $t$ is a summation of every $y$ value from $x=1$ to $ x= 10$. This is an example if $b=1$: $$ t = \sum_{i=1}^{10} y(i) = y(1) + y(2) + y(3) + \dotsb + y(10) = \sum_{i=1}^{10} (2^i-1) $$
How would I find the base $b$ if I had the threshold $t$?
This is how to systematically find a solution:
Your sequence is a non-homogeneous linear recurrence with constant coefficients. From $$ y_{i+1}= 2 y_i + b $$ one can deduce $$ y_{i+2} = 2 y_{i+1} + b $$ so for the difference we have $$ y_{i+2} - y_{i+1} = 2 y_{i+1} + b - (2 y_i + b) = 2 y_{i+1} - 2 y_i $$ or \begin{align} y_{i+2} &= 3 y_{i+1} - 2 y_i \iff \\ y_i &= 3 y_{i-1} - 2 y_{i-2} \end{align} The last equation is a homogeneous linear recurrence with constant coefficients.
It has the characteristic equation $$ \lambda^2 - 3 \lambda + 2 = 0 \iff \\ (\lambda - 3/2)^2 = 9/4 - 2 = 1/4 $$ with roots $\lambda = (3 \pm 1)/2 \in \{ 1, 2 \}$. This gives the solution $$ y_i = k_1 1^i + k_2 2^i $$ Inserting the first two values $y_1 = b$ and $y_2 = 2 y_1 + b = 2 b + b = 3b$ we have the system $$ b = k_1 + 2 k_2 \\ 3b = k_1 + 4 k_2 $$ which gives $2b = 2 k_2$ or $k_2 = b$ and $b = k_1 + 2 b$ or $k_1 = -b$. This gives the solution $$ y_i = b \, 2^i- b = b (2^i - 1) $$
Then your threshold is \begin{align} t &= \sum_{i=1}^{10} y_i \\ &= \sum_{i=1}^{10} b (2^i - 1) \\ &= b \sum_{i=1}^{10} (2^i - 1) \\ &= b \left( \left( \sum_{i=1}^{10} 2^i \right) - \sum_{i=1}^{10} 1 \right) \\ &= b \left( \left( \frac{2^{10+1}-1}{2 - 1} - 1 \right) - 10 \right) \\ &= b ((2047 - 1) - 10) \\ &= 2036 \, b \end{align} where we used the formula for the geometric series with $q=2$, $$ S(q) = \sum_{k=0}^n q^k = \frac{q^{n+1}-1}{q - 1} $$ starting at $k = 1$ instead of $k = 0$.
This gives $$ b = \frac{t}{2036} $$