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  • 1 Post By skipjack
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March 31st, 2016, 07:46 AM   #1
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Geometric progression

I need Geometric progression formula for this (attached)
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Last edited by skipjack; March 31st, 2016 at 07:52 AM.
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March 31st, 2016, 07:52 AM   #2
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$2^{k+1} - 1$
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April 2nd, 2016, 06:28 AM   #3
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You should know, if you are expected to be able to do a problem like this, that the general finite geometric sum (not "progression"- that would be just the sequence 1, 2, 4, ...), $\displaystyle \sum_{i= 0}^N ar^n$, is equal to $\displaystyle \frac{a(1- r^{N+ 1}}{1- r}$. In your example, a= 1, r= 2, and N= k so the sum is $\displaystyle \frac{1- 2^{k+1}}{1- 2}= 2^{k+1}- 1$ as skipjack said.
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