Discover the Derivation of Closed Form Summation | Step-by-Step Explanation

[mohabitar]

How were they able to derive this?

 

[Mark44]

How were they able to derive this?

The sequence in this sum is a geometric sequence.

 

[fzero]

Denote the sum by s_p. Can you find a linear relationship between s_{p+1} and s_p? Now compute s_{p+1} - s_p explicitly and use the relation to solve for s_p.

 

[JonF]

So S_p = 1/2 + 1/2^2 + 1/2^3 + 1/2^4 + 1/2^5+ … + + 1/2^p
What would (1/2)*S_p look like?
What about S_p - (1/2)*S_p?
Now note: S_p - (1/2)*S_p = S_p(1 – ½) = (½)S_p So ask yourself what 2*(½)S_p = S_p looks like, and you should get your desired result

 

[mohabitar]

Sorry the last post confused me. I mean is there a known formula for something like this?

 

[mohabitar]

There is a known formula if this thing starts at 0. The question is how do I make it start at 0 rather than 1. I knew there was a way but I forgot.

 

[fzero]

s_p = \frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2^{p-1}} + \frac{1}{2^p}

\frac{1}{2}s_p = \frac{1}{4} + \frac{1}{8} + \cdots + \frac{1}{2^p} + \frac{1}{2^{p+1}}

s_{p+1} = \frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2^{p-1}} + \frac{1}{2^p}+\frac{1}{2^{p+1}}

What relations between these expressions can you write down?

 

[mohabitar]

None, I really don't see what you did from the 2nd step to the third step.

 

[JonF]

I think subtracting them is easier to see.

S_p = 1/2 + 1/2² + 1/2³ + 1/2⁴ + 1/2⁵ + … + 1/2^p

½S_p = 1/2² + 1/2³ + 1/2⁴ + 1/2⁵ + … + 1/2^p + 1/2^p+1

if we take S_p - ½S_p, which terms would cancel out? Which would be left over? I color coded it to help you see it.

 

[fzero]

None, I really don't see what you did from the 2nd step to the third step.

The 3rd step is just the expansion of the definition

s_{p} = \sum_{k=1}^p \frac{1}{2^k}

applied to p+1.

 

[mohabitar]

Why would we subtract 1/2sp from sp? Can we just make a quick generalization? There was a trick to it that I forgot. If the summation starts at 1 and you want it to start at 0, then just take out the first term, or something like that? But now I still don't get the original question. Any other way to explain-you guys are really complicating this thing.

 

[JonF]

Because it nearly gives us our desired result. Look at how I color coded it. S_p - ½S_p = ½ - 1/2^p+1

 

[fzero]

Why would we subtract 1/2sp from sp? Can we just make a quick generalization? There was a trick to it that I forgot. If the summation starts at 1 and you want it to start at 0, then just take out the first term, or something like that? But now I still don't get the original question. Any other way to explain-you guys are really complicating this thing.

We're telling you how to find the sum without referring to another result. If you know the answer for

S_p = \sum_{k=0}^p \frac{1}{2^p},

then explain where you got it and we'll tell you how to use it here.

 

[mohabitar]

Ok forget about this question. I'm just not seeing it. There should be an equation in the book. But now for the other question, changing the index of the summation, how can I do that? If any summation starts at i=2, and I want to make it start at 0, how would I do that?

 

[JonF]

Don't give up on this problem yet, look at my color coded post (#9), do you see a pattern between the 1/2sp and sp?

Consider this example for changing indexes, actually write out the terms
∑⁵_i=1(i)

∑⁶_i=2(i-1)

 

[HallsofIvy]

The only difference between \sum_{n=0}^\infty r^n and \sum_{n=1}^\infty r^n is that the first is missing the first term, r^0= 1. To find \sum_{n=1}^\infty r^n, just find \sum_{n=0}^\infty r^n and subtract 1.

For example, to find ]\displaytype\sum_{n=1}^\infty \left(\frac{1}{3}\right)^n, first find \displaytype\sum_{n=0}^\infty \left(\frac{1}{3}\right)^n= \frac{1}{1- \frac{1}{3}}= \frac{3}{2}.

Now subtract 1: \frac{3}{2}- 1= \frac{1}{2}

In your original problem, with a finite sum, to find \displaytype\sum_{k=1}^p\frac{1}{2}^p, start from
\displaytype\sum_{k=0}^p \frac{1}{2^p}= \frac{1- \left(\frac{1}{2}\right)^{p+1}}{1- \frac{1}{2}
= 2(1- \frac{1}{2^{p+1}})= 2- \frac{1}{2^p}

Now subtract 1:
\left(2- \frac{1}{2^p}\right)- 1= 1- \frac{1}{2^p}