x+y + xy+x^2+y^2 + x^3+x^2y+xy^2+y^3 + ... n terms

Question

the sum of n terms of the above series is?

Answer 1

The key to this problem is to note that   x^m  - y^m =  ( x - y) ( x^m-1 y⁰ + x^m-2 y¹ + x^m-3 y² + ... + x⁰ y^m-1)

The second term in this product is homogeneous of order m -1  in x and y.

 

By combining n-1 expressions like the one above,   one can show that

 

( x -y)  times   [ ( x + y) + (x² + xy + y² )  + (x³ + x² y + x y² + y³) +   ...    ]  is equal to

[ (x² - y²) + (x³ - y³) + (x⁴ - y⁴)  +   ... ]

 

So the expression in the question is equal to

 

[ x² + x³ + x⁴ + ... + xⁿ - y² - y³ -y⁴ - ... - yⁿ] /(x-y)

 

If n is at least 2   the geometric series sums can be organized as

 

x² (1 - x^n-1) /[x - y) (1 -x) ] - y²(1 - y^n-1) /[(x - y) (1 -y) ]