Question

Write the sum using summation notation, assuming the suggested pattern continues. 4-24+144-864+...

Answer 1

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

∵ $\frac{-24}{4}=-6$

∵ $\frac{144}{-24}=-6$

∴ 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 = $\frac{a(1-r^{n})}{(1-r)}$

∴ Sn = $\frac{4(1-(-6)^{n})}{(1-(-6))}=\frac{4(1-(-6)^{n})}{(1+6)}=\frac{4}{7}[1-(-6)^{n}]$

- By using summation notation

∵ Sn = ∑ a r^n , from n = 0 to n = n

∴ Sn = ∑ 4(-6)^n

Answer 2

Answer:

$a_n = (4)(-6)^{n-1}, n =1,2,3,4,....$

And we can verify:

$n=1 , a_1 = 4 (-6)^{1-1}= 4$

$n=2 , a_2 = 4 (-6)^{2-1}= -24$

$n=3 , a_3 = 4 (-6)^{3-1}= 144$

$n=4 , a_4 = 4 (-6)^{4-1}= -864$

And finally we can write the summation like this:

$S_n = \sum_{i=1}^n 4 (-6)^{n-1} , n =1,2,3,...$

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:

$a_n = a_1 r^{n-1}$

And for this case we can identify the value of r dividing successive terms like this:

$r = \frac{|24|}{|4|}= \frac{|144|}{|24|}=\frac{|864|}{|144|}= 6$

So for this case we know that the value of r =6 and the initial value 4 would represent $a_1 = 4$

Since the sequence is alternating with + and - signs we can express the general term like this:

$a_n = (4)(-6)^{n-1}, n =1,2,3,4,....$

And we can verify:

$n=1 , a_1 = 4 (-6)^{1-1}= 4$

$n=2 , a_2 = 4 (-6)^{2-1}= -24$

$n=3 , a_3 = 4 (-6)^{3-1}= 144$

$n=4 , a_4 = 4 (-6)^{4-1}= -864$

And finally we can write the summation like this:

$S_n = \sum_{i=1}^n 4 (-6)^{n-1} , n =1,2,3,...$