- #1

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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?

- Thread starter mohabitar
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- #1

- 140

- 0

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?

- #2

danago

Gold Member

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The 1/2=2

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?

- #3

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

- #4

danago

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Think about it like this:Oh ok..so how is it that we can change the bound? What is the rule for that or steps?

[tex]\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}} [/tex]

Do you see why changing the bounds is valid?

- #5

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

- #6

danago

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But remember, the index of 2 is now k-1. So when k=n+1, the index will be (n+1)-1=nBut the bound for the first was n+1, so shouldnt the last term be 2^(n+1) or no?

- #7

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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?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

- #8

danago

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Yes thats rightSo 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?

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