# Summation Question

So whats 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?

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danago
Gold Member

So whats 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

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

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

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

Do you see why changing the bounds is valid?

But the bound for the first was n+1, so shouldnt the last term be 2^(n+1) or no?

danago
Gold Member
But the bound for the first was n+1, so shouldnt 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

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
Gold Member
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 thats right