Answer:
[tex]\large\boxed{B.\ \sum\limits_{i=1}^{8}2^{i-1}=255}[/tex]
Step-by-step explanation:
Method 1:
You can calculate the values of the first eight terms and add them.
Put i = 1, i = 2, ..., i = 8 to the expression
[tex]a_i=2^{i-1}[/tex]
[tex]a_1=2^{1-1}=2^0=1\\\\a_2=2^{2-1}=2^1=2\\\\a_3=2^{3-1}=2^2=4\\\\a_4=2^{4-1}=2^3=8\\\\a_5=2^{5-1}=2^4=16\\\\a_6=2^{6-1}=2^5=32\\\\a_7=2^{7-1}=2^6=64\\\\a_8=2^{8-1}=2^7=128\\\\\sum\limits_{i=1}^{8}2^{i-1}=1+2+4+8+16+32+64+128=255[/tex]
Method 2:
The formula of a sum of n terms of a geometric sequence:
[tex]S_i=\dfrac{a_1(1-r^i)}{1-r}[/tex]
We have:
[tex]a_i=2^{i-1}\\\\a_1=2^{1-1}=2^0=1\\\\r=\dfrac{a_{i+1}}{a_i}\\\\a_{i+1}=2^{i+1-1}=2^i\to r=\dfrac{2^i}{2^{i-1}}=2^{i-(i-1)}=2^{i-i+1}=2^1=2\\\\i=8[/tex]
Substitute:
[tex]S_8=\dfrac{1(1-2^8)}{1-2}=\dfrac{1-256}{-1}=\dfrac{-255}{-1}=255[/tex]
