Answer:
![S_n = (1 (1 - 3^(10)))/(1 - 3) = 29524](https://img.qammunity.org/2022/formulas/mathematics/college/frruwbt0os19soyir83a0wtc4q0jyrp1bk.png)
Explanation:
There's a handy formula we can use to find the sum of a geometric sequence, and here it is
![S_n = (a_1 (1 - r^n))/(1 - r)](https://img.qammunity.org/2022/formulas/mathematics/college/uwouhz7d42x57ihv4r5xkdp1ikuphkjl5z.png)
The value n represents the amount of terms you want to sum in the sequence. The variable r is known as the common ratio, and a is just some constant. Let's find those values.
First lets visualize this sequence
![n_1 = 1\\n_2 = 1 + 3\\n_3 = 1 + 3 + 3^2\\n_4=1+3+3^2+3^3\\...](https://img.qammunity.org/2022/formulas/mathematics/college/f0495o1dzkoca7ufp1ltbiq2bftdwdzw5j.png)
Okay so there's clearly a pattern here, let's write it a bit more concisely. For each n, starting at 1, we raise 3 to the (n-1) power, add it to what we had for the previous term.
![S_n = \sum{3^(n-1)} = 3^(1 - 1) + 3^(2 - 1) + 3^(3-1) ...](https://img.qammunity.org/2022/formulas/mathematics/college/brcaph61coeq7vn2c5zpzrzl2jj27ipzu7.png)
Our coefficients of r, and a, are already here! As you can see below, r is just 3, and a is just 1.
![S_n = \sum{a*r^(n-1)}](https://img.qammunity.org/2022/formulas/mathematics/college/jzv63cgheq4nsd7hj7315sxb1ekm0d8s9o.png)
To finish up lets plug these coefficients in and get our sum after 10 terms.
![S_n = (1 (1 - 3^(10)))/(1 - 3) = 29524](https://img.qammunity.org/2022/formulas/mathematics/college/frruwbt0os19soyir83a0wtc4q0jyrp1bk.png)