The Generating Function for the Sequence of Cubes

Problem: Find the generating function for the sequence of cubes:

0, 1, 8, ..., n³, ....

Solution:

Using the formula

(1+x)/(1 -x)³ = 1 + 2²x + 3²x² + ...+ n²x^n-1 +...

we multiply by x:

x(1+x)/(1 -x)³ = x + 2²x² + 3²x³ + ...+ n²xⁿ +...

(x² + x)(1 -x)^-3 = x + 2²x² + 3²x³ + ...+ n²xⁿ +...

Now we differentiate, noting that the right-hand side of the equality will be

1 + 2³x + 3³x² + ...+ n³x^n-1 +...

and the left-hand side will be

(2x + 1)(1 -x)^-3 + 3(x² + x)(1 -x)^-4

= (2x + 1)(1-x)(1 -x)^-4 + 3(x² + x)(1 -x)^-4

= (2x + 1 - 2x² - x + 3x² + 3x)(1 -x)^-4 = (x² + 4x + 1)(1 -x)^-4 = 1 + 2³x + 3³x² + ...+ n³x^n-1 +...

Now we multiply both sides by x and get x(x² + 4x + 1)(1 -x)^-4 = x + 2³x² + 3³x³ + ...+ n³xⁿ +...

Thus, the generating function for the sequence of cubes is x(x² + 4x + 1)(1 -x)^-4

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