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

[tex]a_n=5 \ for \ n=1 \\ a_n=-10 \ for \ n=2 \\ a_n=20 \ for \ n=3[/tex]

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

[tex]a_1=5 \\ r=\frac{a_2}{a_1}=\frac{-10}{5}=-2 \\ \\ a_n=5(-2)^{n-1}[/tex]

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:
[tex]\sum^{15}_{n=4} 5(-2)^{n-1}[/tex]

Love88126mirandid Love88126mirandid · Answer 2

Love88126mirandid Love88126mirandid

Answer:

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

Step-by-step explanation:

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