Question

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

Answer 1

It's a geometric sequence.

$4,-12,36,... \\ \\ a_1=4 \\ r=\frac{a_2}{a_1}=\frac{-12}{4}=-3 \\ \\ a_n=a_1 \times r^{n-1} \\ a_n=4 \times (-3)^{n-1} \\ a_n=4 \times (-3)^{-1} \times (-3)^n \\ a_n=4 \times (-\frac{1}{3}) \times (-3)^n \\ a_n=-\frac{4}{3}(-3)^n$

It's the sum for term 4 through term 15.

 $\boxed{ \sum\limits_{n=4}^{15} (\frac{4}{3}(-3)^n)}$