+285 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.
+129852 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
+285 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 \(n=\boxed{6}\).
