Question

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

Answer 1

Answer:

$\sum^{15} _{n=4} 4(-3)^{n-1}$=14348880

Step-by-step explanation:

Given :

n an

1 4

2     −12

3 36

First term = a = 4

Common ratio = r = $\frac{a_2}{a_1}=\frac{a_3}{a_2}=\frac{-12}{3}=\frac{36}{-12}=-3$

Formula of nth term of G.P. = $a_n=ar^{n-1}$

So, nth term = $a_n=4(-3)^{n-1}$

Now we are supposed to find  the sum for term 4 through term 15.

$\sum^{15} _{n=4} 4(-3)^{n-1}$

So, $14348880$

Hence the sigma notation of the sum for term 4 through term 15 is $\sum^{15} _{n=4} 4(-3)^{n-1}$=14348880

Answer 2

Hi,

Sorry i don't understand you "determine the sigma notation of the sum for term 4 through term 15" 
$a_1=4\\ a_2=-12=4*(-3)\\ a_3=36=(-12)*(-3)=4*(-3)^2$

$a_4=4*(-3)^3\\ a_5=4*(-3)^4\\ ... a_{15}=4*(-3)^{14}\\ \sum_{i=4}^{15}a_i\\ =a_4+a_5+...+a_{15}\\ =4*(-3)^3 + 4*(-3)^4+...+4*(-3)^{14}\\ =4*(-3)^3*(1+(-3)+(-3)^2+(-3)^3+...+(-3)^{11}\\ =4*(-3)^3* \dfrac{(-3)^{12}-1}{(-3-1)} \\ =4*(-27)* \dfrac{3^{12}-1}{(-4)}\\ =27*(531441-1)\\ =14348880$