Why is the sum of exponential function 2^x for n iterations one less than 2^n?

[Bassalisk]

I posted a related thread, in the same forum regarding matlab, but I want to discuss this here separately.

Few minutes ago I became very confused baffled and surprised at the same time.

consider function 2^{x}.

Now consider the sum of this function for n iterations.(from 0 to n)

\sum_{x=0}^n 2^{x}.

Using my humble knowledge of c++ I found that you need 19 iterations (including zero as iteration) for a number 524287.

so 1+2+4+8+16...+262144=524287

By case, found that this sum is 1 less than the 2^{19}.

so 524287=2^{19}-1

Why? I am not best mathematician you will come across but this i found very interesting.

 

[Charles49]

The binomial theorem states that
<br /> (x+y)^n = \sum_{k=0}^n {n \choose k}x^{n-k}y^k = \sum_{k=0}^n {n \choose k}x^{k}y^{n-k}.<br />

Now set, x=1 and y=1.

 

[Bassalisk]

The binomial theorem states that
<br /> (x+y)^n = \sum_{k=0}^n {n \choose k}x^{n-k}y^k = \sum_{k=0}^n {n \choose k}x^{k}y^{n-k}.<br />

Now set, x=1 and y=1.

I don't think I understand.

General rule I got out of my numbers is:

\sum_{i=0}^n 2^{i}=2^{n+1}-1

By putting zeros into my binomial equation i get:

2^n =1+ \sum_{k=1}^n {n \choose k}1^{n-k}1^k=1+\sum_{k=1}^n {n \choose k}

How do I relate factorial formula to \sum_{i=0}^n 2^{i}

 

[AlephZero]

This is an example of a "geometric seriies" (google will find lots of web pages). It is a bit of a "special case" because you used 2 as the multiplying factor.

In general, suppose you have a sum
S = 1 + x + x^2 \cdots + x^{n-1}
You can multiply both sides by x to get
Sx = x + x^2 + x^3 + \cdots + x^n
Now subtract the first sum from the second one. All the terms in the sums cancel out except for the first and last ones, and you get
Sx - S = S(x-1) = x^n -1
When x = 2 the result is even simpler because x-1 = 1, and you get
1 + 2 + 2^2 ... + 2^{n-1} = 2^n - 1

BTW I don't really understand what the Binomial theorem has to do with this either.

 

[Bassalisk]

This is an example of a "geometric seriies" (google will find lots of web pages). It is a bit of a "special case" because you used 2 as the multiplying factor.

In general, suppose you have a sum
S = 1 + x + x^2 \cdots + x^{n-1}
You can multiply both sides by x to get
Sx = x + x^2 + x^3 + \cdots + x^n
Now subtract the first sum from the second one. All the terms in the sums cancel out except for the first and last ones, and you get
Sx - S = S(x-1) = x^n -1
When x = 2 the result is even simpler because x-1 = 1, and you get
1 + 2 + 2^2 ... + 2^{n-1} = 2^n - 1

BTW I don't really understand what the Binomial theorem has to do with this either.

This makes sense. Thank you !