Question on finite and geometric series

In summary, the finite series summed up to 64 is twice the sum of the first 10 terms of the geometric series.
  • #1
Luceian K
7
0
1. 1. Find the exact(no approximations)sum for the finite series

S sub n= (2 + 2 + 2(2+...+64

i used the parentheses to represent a radical sign

2. Show that the sum of the first 10 terms of the geometric series

1 + 1/3 + 1/9 + 1/27+...

is twice the sum of the first 10 terms of the series

1 - 1/3 + 1/9 - 1/27+...




2. no relevant equations to solve this.


3. I attempted to solve this by dividing n/2(a sub 1 +a sub n). n would equal 64 and a sub one would equal radical 2 so 64/2 (radical 2 +64).

This is as far as I have been able to get and am not sure if it is the correct way to solve these two problems. I would appreciate if someone could show me step by step how to go about these since my pre-calc book does horribly at it.
 
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  • #2
Luceian K said:
1. 1. Find the exact(no approximations)sum for the finite series

S sub n= (2 + 2 + 2(2+...+64

i used the parentheses to represent a radical sign
So that is [itex]n= \sqrt{2+ 2+ 2\sqrt{2+ ...+ 64}}[/itex]?
What does the "..." represent?

2. Show that the sum of the first 10 terms of the geometric series

1 + 1/3 + 1/9 + 1/27+...

is twice the sum of the first 10 terms of the series

1 - 1/3 + 1/9 - 1/27+...




2. no relevant equations to solve this.
I would think that
[tex]\sum_{i=0}^n r^i= \frac{1- r^{n+1}}{1- r}[/tex]
would be a very relevant equation!


3. I attempted to solve this by dividing n/2(a sub 1 +a sub n). n would equal 64 and a sub one would equal radical 2 so 64/2 (radical 2 +64).

This is as far as I have been able to get and am not sure if it is the correct way to solve these two problems. I would appreciate if someone could show me step by step how to go about these since my pre-calc book does horribly at it.
 
  • #3
the "+...+64" is not under the radical sign. Just 2)2. and the "..." is in the question, I don't know what it stands for. I am at a complete loss with these two questions. and my math book doesn't go into enough detail with these types of equations.
 
  • #4
Luceian K said:
1. 1. Find the exact(no approximations)sum for the finite series

S sub n= (2 + 2 + 2(2+...+64
Is this [tex]S_n= \sqrt{2} + 2 + 2\sqrt{2}+...+64 = \sum_{1\le k \le 12} 2^{k/2}[/tex] ?

if so, then it's a geometric series. Do you know the closed form of the sum of a geometric series?
If not, here's how to get it

[tex]S = a + ax + ax^2 + ... + ax^n = \sum_{0 \le k \le n} ax^k[/tex]

[tex]Sx = ax + ax^2 + ax^3 + ... + ax^{n+1}[/tex]

[tex]S - Sx = S(1-x) = a- ax^{n+1}[/tex]

so [tex] S = a\frac{1-x^{n+1}}{1-x}[/tex]
 
  • #5
so the equation would not be n/2(a sub 1 + a sub n)
 
  • #6
HallsofIvy said:
So that is [itex]n= \sqrt{2+ 2+ 2\sqrt{2+ ...+ 64}}[/itex]?
What does the "..." represent?


I would think that
[tex]\sum_{i=0}^n r^i= \frac{1- r^{n+1}}{1- r}[/tex]
would be a very relevant equation!

for the second problem could you tell me how to find r i am unsure of how to solve for that variable
 
  • #7
Sorry but I don't understand what you're asking. You don't have to solve for anything, just set [tex]r=\sqrt{2}[/tex] (or [tex]x=\sqrt{2}[/tex] in my post) because [tex]a=1[/tex] in your equation.
 
  • #8
qntty said:
Sorry but I don't understand what you're asking. You don't have to solve for anything, just set [tex]r=\sqrt{2}[/tex] (or [tex]x=\sqrt{2}[/tex] in my post) because [tex]a=1[/tex] in your equation.

then i am not sure what n would be in the equation you have there.
 
  • #9
Luceian K said:
and the "..." is in the question, I don't know what it stands for. I am at a complete loss with these two questions. and my math book doesn't go into enough detail with these types of equations.

Well I guess that's the problem. Ellipses mean that you follow the pattern that has been set until you get to 64 (in this case).

For example [tex]2+4+8+16+...+512[/tex] is shorthand for [tex]2+4+8+16+32+64+128+256+512[/tex].

In your problem, the patter is that the next term in the series is generated by multiplying the previous term by [tex]\sqrt{2}[/tex]
 
  • #10
Luceian K said:
then i am not sure what n would be in the equation you have there.

the final term in a geometric series can be written in the form [tex]ax^n[/tex] for some number n. The equation which you have is

[tex]\sqrt{2} + (\sqrt{2})^2+ (\sqrt{2})^3+ ... + (\sqrt{2})^{12}[/tex]

So n=12
 
  • #11
i got 216.09545 is this correct
 
  • #12
Luceian K said:
i got 216.09545 is this correct

Yes, that's correct. Now try the other problem.
 
  • #13
thank you for your help
 

What is a finite series?

A finite series is a sequence of numbers that has a finite number of terms. It can be expressed using the summation notation, where the terms are added together starting from the first term to the last term.

What is a geometric series?

A geometric series is a series in which each term is obtained by multiplying the previous term by a constant number called the common ratio. It can be expressed using the summation notation, where the first term is multiplied by the sum of the common ratio raised to various powers.

What is the formula for finding the sum of a finite series?

The formula for finding the sum of a finite series is S = (n/2)(a + l), where S is the sum, n is the number of terms, a is the first term, and l is the last term. This formula can be used for both arithmetic and geometric series.

What is the formula for finding the sum of an infinite geometric series?

The formula for finding the sum of an infinite geometric series is S = a / (1 - r), where S is the sum, a is the first term, and r is the common ratio. This formula can only be used if the absolute value of the common ratio is less than 1.

How can we use finite and geometric series in real life?

Finite and geometric series can be used in various real-life situations such as calculating interest rates, population growth, and compound investments. They can also be used in fields like physics and engineering to model and predict natural phenomena.

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