Question 146054
what is the geometric series {{{-9/2}}}+{{{3/2}}}+{{{-1/2}}}+{{{1/6}}}-...+{{{1/39366}}} in summation notation?
<pre><font size = 4 color = "indigo"><b>
{{{sum(a[1]r^(k-1), k=1, N )}}}


We divide the 2nd term by the 1st term to find the common ratio {{{r}}}

{{{(3/2)/((-9)/2)=(3/2)*(2/(-9))=(3/cross(2))*(cross(2)/(-9))=3/(-9)=-1/3}}}

To check we divide the 3rd term by the 2nd term to see if we get the 
same common ratio {{{-1/3}}}:

{{{(-1/2)/(3/2)=(-1/2)(2/3)=(-1/cross(2))(cross(2)/3)=  -1/3}}} 

To double check we divide the 4th term by the 3rd term to see if we get the 
same common ratio {{{-1/3}}}:

{{{(1/6)/(-1/2)=(1/6)*(2/(-1))=2/(-6)=-1/3}}}

Now that we are triple-sure that the common ratio {{{r=-1/3}}},

we will use the formula for the nth term:

{{{a[n] = a[1]r^(n-1)}}}

to find out how many terms it has:

{{{a[n] = 1/39366}}} is the last, or nth, term:
{{{a[1] = -9/2}}} is the first term
{{{r = -1/3}}} is the common ratio.
Substituting:

{{{1/39366=(-9/2)(-1/3)^(n-1)}}}

Multiply both sides by {{{(39366/1)}}}

{{{(39366/1)(1/39366)=(39366/1)(-9/2)(-1/3)^(n-1)}}}
 
{{{1=-177147(-1/3)^(n-1)}}}

Write {{{(-1/3)^(n-1)}}} as {{{((-1)^(n-1)/3^(n-1))}}}

{{{1=-177147((-1)^(n-1)/3^(n-1))}}}

Multiply both sides by {{{3^(n-1)}}}

Observe that {{{177147=3^11}}}, so substitute that:

{{{3^(n-1)=(3^ll)(-1)^(n-1)}}}

Divide both sides by {{{3^11}}}

{{{3^(n-1)/3^11=(-1)^(n-1)}}}

Subtract exponents on the left:

{{{3^(n-1-11)=(-1)^(n-1)}}}

{{{3^(n-12)=(-1)^(n-1)}}}

Since the right side is a power of
{{{-1}}}, it is either 1 or -1

But no power of 3 can be negative, so

{{{3^(n-12)=1}}}

And since the only power of 3 that gives 1
is the 0 power, i.e., 3<sup>0</sup>=1, then
the exponent n-12 must equal 0.

n-12=0
   n=12.

So there are 12 terms.  So 

{{{sum(a[1]r^(k-1), k=1, N )}}}

becomes:

{{{sum((-9/2)(-1/3)^(k-1), k=1, 12 )}}}

write {{{-9/2}}} as {{{((-1)/1)(9/2)}}}

{{{sum(((-1)/1)(9/2)(-1/3)^(k-1), k=1, 12 )}}}

write {{{(-1/3)^(k-1)}}} as {{{(-1)^(k-1)/3^(k-1)}}}

{{{sum(((-1)/1)(9/2)((-1)^(k-1)/3^(k-1)), k=1, 12 )}}}

Write {{{9}}} as {{{3^2}}}:

{{{sum(((-1)/1)(3^2/2)((-1)^(k-1)/3^(k-1)), k=1, 12 )}}}

{{{sum(((-1)3^2(-1)^(k-1))/(2(3^(k-1))), k=1, 12 )}}}

{{{sum(((-1)^1*3^2(-1)^(k-1))/(2(3^(k-1))), k=1, 12 )}}}

Add exponents of {{{-1}}} on top:

{{{sum((3^2(-1)^(k-1+1))/(2(3^(k-1))), k=1, 12 )}}}

{{{sum((3^2(-1)^k)/(2(3^(k-1))), k=1, 12 )}}}

Subtract exponents of 3:

{{{sum(((-1)^k)/(2(3^(k-1-2))), k=1, 12 )}}}

{{{sum(((-1)^k)/(2*3^(k-3)), k=1, 12 )}}}

Edwin</pre>