SOLUTION: Write geometric series -9/2+3/2-1/2+1/6- ••• + 1/39366 in summation notation. Using the formula for the sum of an geometric series, compute the sum in the problem above.

Algebra ->  Sequences-and-series -> SOLUTION: Write geometric series -9/2+3/2-1/2+1/6- ••• + 1/39366 in summation notation. Using the formula for the sum of an geometric series, compute the sum in the problem above.       Log On


   

Question 1131912: Write geometric series -9/2+3/2-1/2+1/6- ••• + 1/39366 in summation notation. Using the formula for the sum of an geometric series, compute the sum in the problem above.
Found 2 solutions by MathLover1, ikleyn:
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
geometric series -9/2+3/2-1/2+1/6- ••• + 1/39366
a%5B1%5D -the first term
r-the "common ratio" between terms
a%5Bn%5D -the nth term

a%5Bn%5D=a%5B1%5D%2Ar%5E%28n-1%29
a%5B1%5D=-9%2F2
r=%283%2F2%29%2F%28-9%2F2%29
r=%281%2F1%29%2F%28-3%2F1%29
r=%28-1%2F3%29
a%5Bn%5D=-%289%2F2%29%2A%28-1%2F3%29%5E%28n-1%29

sum:
S%5Bn%5D=%28a%5B1%5D%281-r%5En%29%29%2F%281-r%29

S%5Bn%5D=%28-%289%2F2%29%281-%28-1%2F3%29%5En%29%29%2F%281-%28-1%2F3%29%29...we need n

use last given term 1%2F39366 and plug it in

a%5Bn%5D=-%289%2F2%29%2A%28-1%2F3%29%5E%28n-1%29...last given term is 1%2F39366

1%2F39366=-%289%2F2%29%2A%28-1%2F3%29%5E%28n-1%29...solve for n

%281%2F39366%29%2F%28-9%2F2%29=%28-1%2F3%29%5E%28n-1%29

-1%2F177147=%28-1%2F3%29%5E%28n-1%29...take log of both sides

-log%281%2F177147%29=-log%28%281%2F3%29%5E%28n-1%29%29

-log%281%2F177147%29=-%28n-1%29log%28%281%2F3%29%29

n-1=%28-log%281%2F177147%29%29%2F%28-log%28%281%2F3%29%29%29

n-1=log%281%2F177147%29%2Flog%28%281%2F3%29%29

n-1=11

n=11%2B1

n=12

=>
S%5B12%5D=%28-%289%2F2%29%281-%28-1%2F3%29%5E12%29%29%2F%281%2B1%2F3%29

S%5B12%5D=%28-%289%2F2%29%281-1%2F531441%29%29%2F%284%2F3%29

S%5B12%5D=%28-%289%2F2%29%28531440%2F531441%29%29%2F%284%2F3%29

S%5B12%5D=%28-265720%2F59049%29%2F%284%2F3%29

S%5B12%5D=-66430%2F19683

S%5B12%5D=-3+7381%2F19683




Answer by ikleyn(52772) About Me  (Show Source):
You can put this solution on YOUR website!
.

            The solution by  @MathLover1  seems to be almost infinitely long;
            but it can be implemented much shorter,  without making unnecessary calculations.
            See my solution below. 


The common ratio is  q = 3%2F2 : -9%2F2 = %28%283%2F2%29%29%2F%28%28-9%2F2%29%29 = 3%2F%28-9%29 = -1%2F3.


The formula for the sum of "n" first term of an geometry progression is


    S = %28a%5B1%5D%2Aq%5En-a%5B1%5D%29%2F%28q-1%29.


We have  a%5B1%5D = -9%2F2  and  a%5B1%5D%2Aq%5E%28n-1%29 = 1%2F39366  (the n-th term).


Therefore,  


    S = %28a%5B1%5D%2Aq%5E%28n-1%29%2Aq+-+a%5B1%5D%29%2F%28q-1%29 = %28%281%2F39366%29%2A%28-1%2F3%29-%28-9%2F2%29%29%2F%28%28-1%2F3%29-1%29 = %281%2F%283%2A39366%29+-+9%2F2%29%2F%28%284%2F3%29%29 = %28%282+-+9%2A3%2A39366%29%29%2F%28%282%2A3%2A39366%29%29%2F%28%284%2F3%29%29 = %282-9%2A3%2A39366%29%2F%288%2A39366%29 = -1062880%2F314928 = -66430%2F19683 = -37381%2F19683.