Question

n=?
1 + 1/2 + 1/4 + ... + 1 / 2^n = 2[1-(1/2)^8]
b1=1
q=1/2

Answer 1

[tex]1+ \frac{1}{2} +\frac{1}{2^2}+...+\frac{1}{2^n}=2[1-( \frac{1}{2} )^8] \\S=1\cdot \frac{( \frac{1}{2} )^{n+1}-1}{ \frac{1}{2} -1} =\\ =2\cdot[1-( \frac{1}{2} )^{n+1}]=2[1-( \frac{1}{2} )^8]\\ n+1=8\\ n=7\\ [/tex]