# Question on finite and geometric series

1. Jun 11, 2009

### Luceian K

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.

2. Jun 11, 2009

### HallsofIvy

So that is $n= \sqrt{2+ 2+ 2\sqrt{2+ ...+ 64}}$?
What does the "..." represent?

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

3. Jun 11, 2009

### Luceian K

the "+...+64" is not under the radical sign. Just 2)2. and the "..." is in the question, I dont know what it stands for. Im at a complete loss with these two questions. and my math book doesnt go into enough detail with these types of equations.

4. Jun 11, 2009

### qntty

Is this $$S_n= \sqrt{2} + 2 + 2\sqrt{2}+...+64 = \sum_{1\le k \le 12} 2^{k/2}$$ ?

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

$$S = a + ax + ax^2 + ... + ax^n = \sum_{0 \le k \le n} ax^k$$

$$Sx = ax + ax^2 + ax^3 + ... + ax^{n+1}$$

$$S - Sx = S(1-x) = a- ax^{n+1}$$

so $$S = a\frac{1-x^{n+1}}{1-x}$$

5. Jun 11, 2009

### Luceian K

so the equation would not be n/2(a sub 1 + a sub n)

6. Jun 11, 2009

### Luceian K

for the second problem could you tell me how to find r i am unsure of how to solve for that variable

7. Jun 11, 2009

### qntty

Sorry but I don't understand what you're asking. You don't have to solve for anything, just set $$r=\sqrt{2}$$ (or $$x=\sqrt{2}$$ in my post) because $$a=1$$ in your equation.

8. Jun 11, 2009

### Luceian K

then i am not sure what n would be in the equation you have there.

9. Jun 11, 2009

### qntty

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 $$2+4+8+16+...+512$$ is shorthand for $$2+4+8+16+32+64+128+256+512$$.

In your problem, the patter is that the next term in the series is generated by multiplying the previous term by $$\sqrt{2}$$

10. Jun 11, 2009

### qntty

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

$$\sqrt{2} + (\sqrt{2})^2+ (\sqrt{2})^3+ ... + (\sqrt{2})^{12}$$

So n=12

11. Jun 11, 2009

### Luceian K

i got 216.09545 is this correct

12. Jun 11, 2009

### qntty

Yes, that's correct. Now try the other problem.

13. Jun 11, 2009