Question

Can someone please help me with this question?

Answer 1

`\large\boxed{B.\ \sum\limits_(i=1)^(8)2^(i-1)=255}`

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

`a_i=2^(i-1)`

`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`

Method 2:

The formula of a sum of n terms of a geometric sequence:

`S_i=(a_1(1-r^i))/(1-r)`

We have:

`a_i=2^(i-1)\\\\a_1=2^(1-1)=2^0=1\\\\r=(a_(i+1))/(a_i)\\\\a_(i+1)=2^(i+1-1)=2^i\to r=(2^i)/(2^(i-1))=2^(i-(i-1))=2^(i-i+1)=2^1=2\\\\i=8`

Substitute:

`S_8=(1(1-2^8))/(1-2)=(1-256)/(-1)=(-255)/(-1)=255`