Find the summation from i = 0 to n of 2^i.

Any links that are provided is appreciated

Question

Find the summation from i = 0 to n of 2^i. 

Any links that are provided is appreciated

experimentx · Answer

what's the formula of geometric sequence sum?

anonymous · Answer

(1-r^n)/1-r. This is not a fraction, is it still geometric series?

experimentx · Answer

use it ... the this is geometric series. with r=2, a = 1

experimentx · Answer

just change this into 
(r^n-1)/r-1

anonymous · Answer

So then the answer is 1-2^n / 1 - 2 = -(1-2^n)?

experimentx · Answer

experimentx · Answer

including i=zero, you have n+1 terms.

anonymous · Answer

ok

experimentx · Answer

experimentx · Answer

anonymous · Answer

Thanks, I'm aware of the formula, I just don' t see how wolfram alpha gets their answer

experimentx · Answer

I get's it's anwer by adding!! it's just a computer.

anonymous · Answer

There is no final answer since the answer contains n

experimentx · Answer

W|A doesn't give you answer in that case.
your answer is simply
$$ 2^{n+1} - 1 $$

anonymous · Answer

Right, and if I plug that into the gemoetric series that is not what I get

experimentx · Answer

Yeah ..

anonymous · Answer

So the answer varies with what wolfram alpha yields

experimentx · Answer

No ... just WA does numerical calculation for you.

anonymous · Answer

I don't care about the numerical calculation

experimentx · Answer

anonymous · Answer

When I to it myself I get 1-(2^n)/-1
That is for when I plug it into the formula. 
WA differs obviously

experimentx · Answer

change that n into n+1

anonymous · Answer

Why?

experimentx · Answer

see the formula ... that formula is for n terms ... you have n+1 terms.

anonymous · Answer

The summation is from i = 0 to n. I do not see a n+1

anonymous · Answer

Unless you said that backwards

experimentx · Answer

1,2, ....,n <-- these are n terms 
include 0 <-- this makes n+1 terms.

anonymous · Answer

WA has a positive result, mine is negative

experimentx · Answer

1-(2^n)/-1 = -1(-(2^n - 1)) ... just multiply it by -1 on both numerator and denominator.

anonymous · Answer

What let's you multiply by a -1 on the numerator and denominator? Don't you need this to be equal to something in order to do that?

experimentx · Answer

no .. you can always do that.

anonymous · Answer

hm. ok, thank you

experimentx · Answer

yw ... and change that n to n+1 if you are summing from i=0 to i=n

anonymous · Answer

Got it