Summation Question: Subtracting 1, 2^(k-1) & 1/2

[mohabitar]

So what's going on here? Since there is a 2^(k-1), I can subtract one from n and also the index? Thats what it looks like they did. Also, where did they get that 1/2 from?

 

[danago]

So what's going on here? Since there is a 2^(k-1), I can subtract one from n and also the index? Thats what it looks like they did. Also, where did they get that 1/2 from?

The 1/2=2^-1 is simply the term of the summation for which k=-1; by bringing it out of the summation, the lower bound can be changed from k=-1 to k=0

 

[mohabitar]

Oh ok..so how is it that we can change the bound? What is the rule for that or steps?

 

[danago]

Oh ok..so how is it that we can change the bound? What is the rule for that or steps?

Think about it like this:

\sum\limits_{k = 0}^{n + 1} {{2^{k - 1}}} = {2^{ - 1}} + {2^0} + {2^1} + ... + {2^{n - 1}} + {2^n} = \sum\limits_{k = - 1}^n {{2^k}}

Do you see why changing the bounds is valid?

 

[mohabitar]

But the bound for the first was n+1, so shouldn't the last term be 2^(n+1) or no?

 

[danago]

But the bound for the first was n+1, so shouldn't the last term be 2^(n+1) or no?

But remember, the index of 2 is now k-1. So when k=n+1, the index will be (n+1)-1=n

 

[mohabitar]

The 1/2=2-1 is simply the term of the summation for which k=-1; by bringing it out of the summation, the lower bound can be changed from k=-1 to k=0

So if my lower bound were 1, I could bring out a 2 and the new index would start from 2? And if the bound were 2, I could bring out a 4 and new bound would be 3?

 

[danago]

So if my lower bound were 1, I could bring out a 2 and the new index would start from 2? And if the bound were 2, I could bring out a 4 and new bound would be 3?

Yes that's right