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The sum of the first n terms in the infinite geometric sequence \(\left\{\frac{1}{4},\frac{1}{8},\frac{1}{16},\dots \right\}\) is \(\frac{63}{128}\). Find n.

 Jan 5, 2021
 #1
avatar+114092 
+2

We can work backwards here....the common ratio  is  1/8  / 1/4   =   4/8 =  1/2

 

63/ 128  =   (1/4)  ( 1 - (1/2)^n )  / ( 1 - 1/2)

 

63/128  =  (1/4) ( 1 - (1/2)^n) / (1/2)

 

63/128 =  (1/4) (2/1)  ( 1 - (1/2)^n)

 

63/128 =  (1/2) ( 1 - (1/2)^n)

 

(2/1) (63/128) =   1  - (1/2)^n

 

63/ 64  =  1 - (1/2)^n

 

(1/2)^n  =   1 - 63/64

 

(1/2)^n =   64/64 - 63/64

 

1/2^n =  1/64

 

2^n =   64

 

n =   6

 

 

cool cool cool

 Jan 5, 2021
 #2
avatar+312 
+2

Thx CPhill I got it this way 

This is a geometric sequence with first term 1/4 and common ratio 1/2. Thus the sum of the first n terms is: 

\(\frac{63}{128}=\frac{1}{4}\left(\frac{1-\left(\frac{1}{2}\right)^n}{1-\frac{1}{2}}\right)=\frac{2^n-1}{2^{n+1}}\).

We see that \(\frac{63}{128}=\frac{2^6-1}{2^7}\), so \(n=\boxed{6}\).

 Jan 5, 2021
 #3
avatar+114092 
0

Yeah....that's  a good way, as welll!!!!

 

 

cool cool cool

CPhill  Jan 5, 2021

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