Respuesta :
Answer:
Sn = ∑ 4(-6)^n, from n = 0 to n = n
Step-by-step explanation:
* Lets study the geometric pattern
- There is a constant ratio between each two consecutive numbers
- Ex:
# 5 , 10 , 20 , 40 , 80 , ………………………. (×2)
# 5000 , 1000 , 200 , 40 , …………………………(÷5)
- The sum of n terms is Sn = [tex]\frac{a(1-r^{n})}{(1-r)}[/tex], where
a is the first term , r is the common ratio between each two
consecutive terms and n is the numbers of terms
- The summation notation is ∑ a r^n, from n = 0 to n = n
* Now lets solve the problem
∵ The terms if the sequence are:
4 , -24 , 144 , -864 , ........
∵ [tex]\frac{-24}{4}=-6[/tex]
∵ [tex]\frac{144}{-24}=-6[/tex]
∴ There is a constant ratio between each two consecutive terms
∴ The pattern is geometric
- The first term is a
∴ a = 4
- The constant ratio is r
∴ r = -6
∵ Sn = [tex]\frac{a(1-r^{n})}{(1-r)}[/tex]
∴ Sn = [tex]\frac{4(1-(-6)^{n})}{(1-(-6))}=\frac{4(1-(-6)^{n})}{(1+6)}=\frac{4}{7}[1-(-6)^{n}][/tex]
- By using summation notation
∵ Sn = ∑ a r^n , from n = 0 to n = n
∴ Sn = ∑ 4(-6)^n
Answer:
[tex] a_n = (4)(-6)^{n-1}, n =1,2,3,4,.... [/tex]
And we can verify:
[tex] n=1 , a_1 = 4 (-6)^{1-1}= 4[/tex]
[tex] n=2 , a_2 = 4 (-6)^{2-1}= -24[/tex]
[tex] n=3 , a_3 = 4 (-6)^{3-1}= 144[/tex]
[tex] n=4 , a_4 = 4 (-6)^{4-1}= -864[/tex]
And finally we can write the summation like this:
[tex] S_n = \sum_{i=1}^n 4 (-6)^{n-1} , n =1,2,3,... [/tex]
Step-by-step explanation:
For this case we have the following pattern of numbers :
4-24+144-864+...
And we want to express the sum in terms of a summation.
We can use the fact the the general term for the sum can be expressed as:
[tex] a_n = a_1 r^{n-1}[/tex]
And for this case we can identify the value of r dividing successive terms like this:
[tex] r = \frac{|24|}{|4|}= \frac{|144|}{|24|}=\frac{|864|}{|144|}= 6[/tex]
So for this case we know that the value of r =6 and the initial value 4 would represent [tex] a_1 = 4[/tex]
Since the sequence is alternating with + and - signs we can express the general term like this:
[tex] a_n = (4)(-6)^{n-1}, n =1,2,3,4,.... [/tex]
And we can verify:
[tex] n=1 , a_1 = 4 (-6)^{1-1}= 4[/tex]
[tex] n=2 , a_2 = 4 (-6)^{2-1}= -24[/tex]
[tex] n=3 , a_3 = 4 (-6)^{3-1}= 144[/tex]
[tex] n=4 , a_4 = 4 (-6)^{4-1}= -864[/tex]
And finally we can write the summation like this:
[tex] S_n = \sum_{i=1}^n 4 (-6)^{n-1} , n =1,2,3,... [/tex]