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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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# Recurrence Relation help

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