Question

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

n an
1 4
2 −12
3 36

Answer 1

Hi,

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

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

Answer 2

The sum of the term 4 through term 15 is 14,348,880

Given the sequence of numbers 4, -12, 36...

The sequence is geometric in nature.

The sum of the nth term of the sequence will be given as:

$S_n=\frac{a(1-r^n)}{1-r}$

• a is the first term
• r is the common ratio
• n is the number of terms

For this sequence, the 4 term will serve as our "a"

a = T4 = (36 * -3) = -108

r = -12/4 = 36/-12 = -3

n = 12 (4th to 15term)

Substitute into the formula;

$S_{12} = \frac{-108(1-(-3)^{12}}{1+3}\\ S_{12} = \frac{-108(1-(-3)^{12})}{1+3}\\S_{12} = \frac{-108(531,440)}{4}\\S_{12} = 14,348,880$

Hence the sum of the term 4 through term 15 is 14,348,880

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