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seansrk
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[PLAIN]http://img442.imageshack.us/img442/4514/ballsp.jpg
The base case is 2[itex]^{K}[/itex] = n (which turns into log[itex]_{2}[/itex]n = k
So I have a question on this recurrence relation problem. (I'm trying to get to make the top equation look like the bottom equation.) I know that the summation ends up becoming
[itex]\frac{((7/4)^K)-1}{(7/4)-1}[/itex]
which gives us
((7/4)^log[itex]_{2}[/itex]n)-1 / (3/4)
I'm lost on how the n drops down. I know how the (log[2]7-2)-1 comes to be but how does the n and log[2]7 -2 get on the same level and the exponent level disappeared?
Hopefully I didn't make this confusing. Just trying to find out how the n came down.
The base case is 2[itex]^{K}[/itex] = n (which turns into log[itex]_{2}[/itex]n = k
So I have a question on this recurrence relation problem. (I'm trying to get to make the top equation look like the bottom equation.) I know that the summation ends up becoming
[itex]\frac{((7/4)^K)-1}{(7/4)-1}[/itex]
which gives us
((7/4)^log[itex]_{2}[/itex]n)-1 / (3/4)
I'm lost on how the n drops down. I know how the (log[2]7-2)-1 comes to be but how does the n and log[2]7 -2 get on the same level and the exponent level disappeared?
Hopefully I didn't make this confusing. Just trying to find out how the n came down.
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