Question 73789
Starting with the sequence of: 1, 1/2, 1/4, 1/8,...
a) Find the common ratio.
The common ratio, r, is found by dividing any term by the preceeding term.
{{{r = (1/2)/1}}}
{{{r = 1/2}}}
or
{{{r = (1/8)/(1/4)}}} 
{{{r = (1/8)*(4/1)}}}
{{{r = 1/2}}}
The common ratio is 1/2

b) The partial sum of the first n terms of a geometric series is given by:
{{{S[n] = (a[1](1-r^n))/(1-r)}}} where:
{{{n}}} is the number of the term (1st, 2nd, 3rd,...)
{{{a[1]}}} is the first term.
{{{r}}} is the common ratio. 
To find the partial sum of the first 10 terms, set {{{n = 10}}}, {{{a[1] = 1}}}, {{{r = 1/2}}}.  Substitute these values into the formula for the partial sum. 
{{{S[10] = (1(1-(1/2)^10))/(1-(1/2))}}}
{{{S[10] = 0.9990/0.5}}}
{{{S[10] = 1.9980}}} To four decimal places.
You should be able to finish the other parts using the above as a guide.
If you have trouble with it, please re-post.

c) Find the partial sum of the first 12 terms.
For this part, n = 12 and we can use the same formula for the partial sum of the first n terms of a geometric sequence.
{{{S[12] = (1(1-(1/2)^12))/(1-1/2)}}}
{{{S[12] = 0.9998/(1/2)}}}
{{{S[12] = 1.9995}}} To four deicimal places.

d) My observation is that the sum approaches 2.
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