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avatar+1666 

Find the sum of the first 9 terms in the following geometric series.
Do not round your answer.

64+32+16+...

I'm writing this one out but I got confused after:

64+32+16+8+4+2+???????????

 

 

EDIT:

I kind of figured it out....

is it 64+32+16+8+4+2+1+1/2+1/4?

then its:

127.75 right?

 Aug 6, 2019
edited by tommarvoloriddle  Aug 6, 2019
 #1
avatar+6045 
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\(\text{I guess the series they are after has $a=64,~r=\dfrac 1 2$}\\ 64 \sum \limits_{k=0}^8\left(\dfrac 1 2 \right)^k = 64 \dfrac{1-\left(\frac 1 2 \right)^9}{1 - \frac 1 2} = \\ 64 \dfrac{1 - \frac{1}{512}}{\frac 1 2} = \dfrac{511}{4}=127.75\\ \text{Well done!}\)

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 Aug 6, 2019
 #2
avatar+1666 
+2

Thank you Rom!

tommarvoloriddle  Aug 6, 2019

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