Question

Given the sequence in the table below, determine the sigma notation of the sum for term 4 through term 15

Answer 1

$a_n=5 \ for \ n=1 \\ a_n=-10 \ for \ n=2 \\ a_n=20 \ for \ n=3$

You can see it's a geometric sequence because the ratio is constant ($\frac{a_3}{a_2}=\frac{a_2}{a_1}$).
The nth term of a geometric sequence:
$a_n=a_1 \times r^{n-1}$

$a_1=5 \\ r=\frac{a_2}{a_1}=\frac{-10}{5}=-2 \\ \\ a_n=5(-2)^{n-1}$

It has to be the sum for term 4 through term 15, so n=4 and the number above the sigma is 15.

Your answer:
$\sum^{15}_{n=4} 5(-2)^{n-1}$

Answer 2

Answer:

the summation of 4 times negative 3 to the n minus 1 power, from n equals 4 to 15.

Step-by-step explanation: