Answer:
![\sum_(n = 1) 2*(-5)^(n-1)](https://img.qammunity.org/2022/formulas/mathematics/college/hsmvppogdow3ivnqr7clbspswkn0dmp0g6.png)
Explanation:
An arithmetic sequence is of the form:
![A_n = A_(n-1) + d](https://img.qammunity.org/2022/formulas/mathematics/college/9q5ijpht9u3jm7zb550fc4js0brv2yg501.png)
While a geometric sequence is of the form:
![A_n = A_(n-1)*r](https://img.qammunity.org/2022/formulas/mathematics/college/u9704rz62v1jh3mxt884fbdih2abq91tfp.png)
notice that first, we have a change of sign in our sequence, so we already can discard the arithmetic sequence.
In fact, the pattern is kinda easy to see.
The first term is:
A₁ = 2
the second term is:
A₂ = -10
notice that:
A₂/A₍ = r = -10/2 = -5
The third term is:
A₃ = 50
the quotient between the third term and the second term is:
A₃/A₂ = 50/-10 = -5
Whit this we can already conclude that the n-th term of our sequence will be:
![A_n = A_(n-1)*(-5)](https://img.qammunity.org/2022/formulas/mathematics/college/tpsxcfioj3opbo3m1p581jwbgta3kswq9u.png)
Then the summation will be something like:
![\sum_(n = 1) A_n = A_1 + A_2 + A_3 + ... = 2 - 10 + 50 - ...](https://img.qammunity.org/2022/formulas/mathematics/college/kgq6m4zvqqbhk7w6g4g5gbio2go5xygro3.png)
We can write:
![A_n = A_(n-1)*(-5) = (A_(n-2)*(-5))*(-5)) = A_1*(-5)^(n-1) = 2*(-5)^(n-1)](https://img.qammunity.org/2022/formulas/mathematics/college/rsnq2srqj2l8jvnb8yo3dtp0sgrw6sq7at.png)
Then the summation is just:
![\sum_(n = 1) 2*(-5)^(n-1)](https://img.qammunity.org/2022/formulas/mathematics/college/hsmvppogdow3ivnqr7clbspswkn0dmp0g6.png)