Answer:
[tex]\sum_{n = 1} 2*(-5)^{n-1}[/tex]
Step-by-step explanation:
An arithmetic sequence is of the form:
[tex]A_n = A_{n-1} + d[/tex]
While a geometric sequence is of the form:
[tex]A_n = A_{n-1}*r[/tex]
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:
[tex]A_n = A_{n-1}*(-5)[/tex]
Then the summation will be something like:
[tex]\sum_{n = 1} A_n = A_1 + A_2 + A_3 + ... = 2 - 10 + 50 - ...[/tex]
We can write:
[tex]A_n = A_{n-1}*(-5) = (A_{n-2}*(-5))*(-5)) = A_1*(-5)^{n-1} = 2*(-5)^{n-1}[/tex]
Then the summation is just:
[tex]\sum_{n = 1} 2*(-5)^{n-1}[/tex]
