Question 1043793
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Let me reformulate the problem in this way:


<pre>
   x and y are real numbers such that |x| < 1  and |y|< 1.
   Find the infinite sum  {{{x+y +(x^2+xy+y^2)+(x^3+x^2y+xy^2+y^3)+ ellipsis}}}
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<U>Solution</U>


<pre>
Let "S" be the infinite sum 

S = {{{x+y +(x^2+xy+y^2)+(x^3+x^2y+xy^2+y^3)+ ellipsis}}}.

Multiply S by (x-y). Then

S = {{{(x+y)*(x-y) + (x^2+xy+y^2)*(x-y) + (x^3+x^2y+xy^2+y^3)*(x-y) + ellipsis}}}.

Now notice that

{{{(x+y)*(x-y)}}} = {{{x^2 - y^2}}},               (1)

{{{(x^2+xy+y^2)*(x-y)}}} = {{{x^3 - y^3}}},        (2)

{{{(x^3+x^2y+xy^2+y^3)*(x-y)}}} = {{{x^4 - y^4}}}  (3)   (make yourself this calc . . . )

And so on . . . 

So, I suppose (and I am almost sure) that each parenthesed term in the original sum, multiplied by (x-y) will give {{{x^n - y^n}}}.   (4)

   //"The margins of this page are too narrow . . . "

Thus we have

  S*(x-y) = {{{(x^2 + x^3 + x^4 + ellipsis)}}} - {{{(y^2 + y^3 + y^4 + ellipsis)}}} =     (5)

     Now apply the formula for the infinite sum of a geometric progression

= {{{x^2/(1-x) - y^2/(1-y)}}}.

Simplify it and then cancel the factor (x-y) in both sides.

Finally, you will get


    S = {{{(x - xy + y)/((1-x)*(1-y))}}} = 1 - {{{1/((1-x)*(1-y))}}}.


Again, the key is the idea with the formulas (1), (2), (3), (4), (5).

<U>Answer</U>.  S = {{{(x - xy + y)/((1-x)*(1-y))}}} = 1 - {{{1/((1-x)*(1-y))}}}.
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