SOLUTION: (x-(-y))-(-(x^2-(-xy)-(-y^2)))-(-(x^3-(-x^2y)-(-xy^2)-(-y^3)))-......... find the sum of n terms of this series.

Algebra ->  Sequences-and-series -> SOLUTION: (x-(-y))-(-(x^2-(-xy)-(-y^2)))-(-(x^3-(-x^2y)-(-xy^2)-(-y^3)))-......... find the sum of n terms of this series.       Log On


   



Question 1012868: (x-(-y))-(-(x^2-(-xy)-(-y^2)))-(-(x^3-(-x^2y)-(-xy^2)-(-y^3)))-.........
find the sum of n terms of this series.

Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
It's easier to see when you write with larger parentheses enclosing
smaller parentheses like this:




First we remove the smallest parentheses: 



Remove the medium-sized parentheses:




Remove the largest parentheses and the remaining terms will also
have positive coefficients, so we put + before the three dots:



Now that all terms have positive coefficients,
let's put parentheses back to separate the terms, and call it
expression (1)

(1)    

Now we use the fact that x%5Ek-y%5Ek factors as 

x%5Ek-y%5Ek%22%22=%22%22

Dividing both sides by (x-y)

%28x%5Ek-y%5Ek%29%2F%28x-y%29%22%22=%22%22

Using that, expression (1) becomes:



And now we can conveniently write the final nth term. 
Notice that the exponent in the nth term will be n+1, because the
exponent in the 1st term is 2, the exponent in the 2nd term is 3,
etc.




Factor out 1%2F%28x-y%29



Remove the inner parentheses:



Rearranging the terms:



Put the x terms in parentheses and write the y terms in parentheses 
preceded by a - sign:

(2)    

Now we look at the sequences

x%5E2%2Bx%5E3%2Bx%5E4%2B%22%22%2A%22%22%2A%22%22%2A%22%22%2Bx%5E%28n%2B1%29 and y%5E2%2By%5E3%2By%5E4%2B%22%22%2A%22%22%2A%22%22%2A%22%22%2By%5E%28n%2B1%29

These are geometric series with n terms, so we use the formula

S%5Bn%5D%22%22=%22%22a%5Bn%5D%28r%5En-1%29%2F%28r-1%29

with x%5E2%2Bx%5E3%2Bx%5E4%2B%22%22%2A%22%22%2A%22%22%2A%22%22%2Bx%5E%28n%2B1%29, r=x and a%5B1%5D=x%5E2

x%5E2%2Bx%5E3%2Bx%5E4%2B%22%22%2A%22%22%2A%22%22%2A%22%22%2Bx%5E%28n%2B1%29%22%22=%22%22%28x%5E2%28x%5En-1%29%29%2F%28x-1%29

Similarly,

y%5E2%2By%5E3%2By%5E4%2B%22%22%2A%22%22%2A%22%22%2A%22%22%2By%5E%28n%2B1%29%22%22=%22%22%28y%5E2%28y%5En-1%29%29%2F%28y-1%29

Substituting those in expression (2),



I'm not going to type out all the algebra, but this can be written



Edwin